Let a common tangent to the curves ${y^2} = 4x$ and ${(x - 4)^2} + {y^2} = 16$ touch the curves at the points P and Q. Then ${(PQ)^2}$ is equal to __________.
Explanation:
$y^2=4 x$ is given by $y=m x+\frac{1}{m}$ and
Tangent of slope $m$ to the circle $(x-4)^2+y^2=16$ is given by
$ y=m(x-4) \pm 4 \sqrt{1+m^2} $
For common tangent
$ \begin{aligned} & \frac{1}{m}=-4 m \pm 4 \sqrt{1+m^2} \\\\ & \Rightarrow \left(\frac{1}{m}+4 m\right)^2=\left( \pm 4 \sqrt{1+m^2}\right)^2 \end{aligned} $
On squaring both sides, we get
$ \begin{aligned} & =\frac{1}{m^2}+16 m^2+8=16+16 m^2 \\\\ & \Rightarrow \frac{1}{m^2} =8 \Rightarrow m= \pm \frac{1}{2 \sqrt{2}} \end{aligned} $
Then, the point of contact on parabola is $(8,4 \sqrt{2})$
Length of tangent $P Q$ from $(8,4 \sqrt{2})$ on the circle is
$ \begin{array}{ll} &\Rightarrow P Q=\sqrt{(8-4)^2+(4 \sqrt{2})^2-16} \\\\ &\Rightarrow P Q=\sqrt{16+32-16} \\\\ &\Rightarrow P Q=\sqrt{32} \Rightarrow(P Q)^2=32 \end{array} $
The ordinates of the points P and $\mathrm{Q}$ on the parabola with focus $(3,0)$ and directrix $x=-3$ are in the ratio $3: 1$. If $\mathrm{R}(\alpha, \beta)$ is the point of intersection of the tangents to the parabola at $\mathrm{P}$ and $\mathrm{Q}$, then $\frac{\beta^{2}}{\alpha}$ is equal to _______________.
Explanation:
Let the tangent to the curve $x^{2}+2 x-4 y+9=0$ at the point $\mathrm{P}(1,3)$ on it meet the $y$-axis at $\mathrm{A}$. Let the line passing through $\mathrm{P}$ and parallel to the line $x-3 y=6$ meet the parabola $y^{2}=4 x$ at $\mathrm{B}$. If $\mathrm{B}$ lies on the line $2 x-3 y=8$, then $(\mathrm{AB})^{2}$ is equal to ___________.
Explanation:
$x^2+2 x-4 y+9=0$ ..........(i)
Equation of tangent at $P(1,3)$ to the given curve (i)
$ \begin{array}{rlrl} & x(1)+2\left(\frac{x+1}{2}\right)-4\left(\frac{y+3}{2}\right)+9 =0 \\\\ & \Rightarrow 2 x+2 x+2-4 y-12+18 =0 \\\\ &\Rightarrow 4 x-4 y+8 =0 \\\\ &\Rightarrow x-y+2 =0 \end{array} $
which is meet the $Y$-axis at $A$
$\therefore A \equiv(0,2)$
Equation of line passing through $P$ and parallel to $x-3 y=6$ is $x-3 y+8=0$
Since, line (ii) meet the parabola $y^2=4 x$ at $B$
$ \begin{array}{lc} &\therefore y^2=4(3 y-8) \\\\ &\Rightarrow y^2-12 y+32=0 \\\\ &\Rightarrow (y-4)(y-8)=0 \end{array} $
$\therefore$ Possible co-ordinates of $B$ are $(4,4)$ and $(16,8)$.
Since, $(4,4)$ does not satisfies line $2 x-3 y=8$
Thus, $B$ is $(16,8)$
$ \therefore (A B)^2=(16-0)^2+(8-2)^2=256+36=292 $
If the $x$-intercept of a focal chord of the parabola $y^{2}=8x+4y+4$ is 3, then the length of this chord is equal to ____________.
Explanation:
Put $(3,0)$ in the above line $\mathrm{m}=-1$
Length of focal chord $=16$
Explanation:
As $\mathrm{P}(\mathrm{b}, \mathrm{c})$ lies on parabola so $\mathrm{c}^{2}=2 \mathrm{ab}---(1)$
Now equation of tangent to parabola $\mathrm{y}^{2}=2 \mathrm{ax}$ in point form is
$\mathrm{yy}_{1}=2 \mathrm{a} \frac{\left(\mathrm{x}+\mathrm{x}_{1}\right)}{2},$
Here, $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(\mathrm{b}, \mathrm{c})$
$\Rightarrow \mathrm{yc}=\mathrm{a}(\mathrm{x}+\mathrm{b})$
For point $\mathrm{B}$, put $\mathrm{y}=0$, now $\mathrm{x}=-\mathrm{b}$
So, area of $\triangle \mathrm{PBA}, \frac{1}{2} \times \mathrm{AB} \times \mathrm{AP}=16$
$ \begin{aligned} & \Rightarrow \frac{1}{2} \times 2 b \times c=16 \\\\ & \Rightarrow b c=16 \end{aligned} $
As $\mathrm{b}$ and $\mathrm{c}$ are natural number so possible values of $(b, c)$ are $(1,16),(2,8),(4,4),(8,2)$ and $(16,1)$
Now from equation (1) $\mathrm{a}=\frac{\mathrm{c}^2}{2 \mathrm{~b}}$ and $\mathrm{a} \in \mathrm{N}$, so values of $(b, c)$ are $(1,16),(2,8)$ and $(4,4)$ now values of are 128,16 and 2 .
Hence sum of values of $a$ is 146 .
A triangle is formed by the tangents at the point (2, 2) on the curves $y^2=2x$ and $x^2+y^2=4x$, and the line $x+y+2=0$. If $r$ is the radius of its circumcircle, then $r^2$ is equal to ___________.
Explanation:
Tangent for ${y^2} = 2x$ at (2, 2) is
${L_1}:2y = x + 2$
Tangent for ${x^2} + {y^2} = 4x$ at (2, 2) is
${L_2}:y = 2$
${L_3}:x + y = 2 = 0$
Radius of circumcircle $ = {{abc} \over {4\Delta }}$
$ = {{(\sqrt {20} )(6)(\sqrt 8 )} \over {4 \times {1 \over 2} \times 6 \times 2}}$
$R = \sqrt {10} $
$R^2=10$
Let the focal chord of the parabola $\mathrm{P}: y^{2}=4 x$ along the line $\mathrm{L}: y=\mathrm{m} x+\mathrm{c}, \mathrm{m}>0$ meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola $\mathrm{H}: x^{2}-y^{2}=4$. If O is the vertex of P and F is the focus of H on the positive x-axis, then the area of the quadrilateral OMFN is :
If the tangents drawn at the points $\mathrm{P}$ and $\mathrm{Q}$ on the parabola $y^{2}=2 x-3$ intersect at the point $R(0,1)$, then the orthocentre of the triangle $P Q R$ is :
If the length of the latus rectum of a parabola, whose focus is $(a, a)$ and the tangent at its vertex is $x+y=a$, is 16, then $|a|$ is equal to :
Let $P(a, b)$ be a point on the parabola $y^{2}=8 x$ such that the tangent at $P$ passes through the centre of the circle $x^{2}+y^{2}-10 x-14 y+65=0$. Let $A$ be the product of all possible values of $a$ and $B$ be the product of all possible values of $b$. Then the value of $A+B$ is equal to :
Let $\mathrm{P}$ and $\mathrm{Q}$ be any points on the curves $(x-1)^{2}+(y+1)^{2}=1$ and $y=x^{2}$, respectively. The distance between $P$ and $Q$ is minimum for some value of the abscissa of $P$ in the interval :
The equation of a common tangent to the parabolas $y=x^{2}$ and $y=-(x-2)^{2}$ is
The tangents at the points $A(1,3)$ and $B(1,-1)$ on the parabola $y^{2}-2 x-2 y=1$ meet at the point $P$. Then the area (in unit ${ }^{2}$ ) of the triangle $P A B$ is :
Let P : y2 = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of ${\pi \over 4}$ with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is :
Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of ${\pi \over 2}$ at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse $E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$, ${a^2} > {b^2}$. If e is the eccentricity of the ellipse E, then the value of ${1 \over {{e^2}}}$ is equal to :
If vertex of a parabola is (2, $-$1) and the equation of its directrix is 4x $-$ 3y = 21, then the length of its latus rectum is :
If the equation of the parabola, whose vertex is at (5, 4) and the directrix is $3x + y - 29 = 0$, is ${x^2} + a{y^2} + bxy + cx + dy + k = 0$, then $a + b + c + d + k$ is equal to :
Let the normal at the point on the parabola y2 = 6x pass through the point (5, $-$8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :
If the line $y = 4 + kx,\,k > 0$, is the tangent to the parabola $y = x - {x^2}$ at the point P and V is the vertex of the parabola, then the slope of the line through P and V is :
If $y = {m_1}x + {c_1}$ and $y = {m_2}x + {c_2}$, ${m_1} \ne {m_2}$ are two common tangents of circle ${x^2} + {y^2} = 2$ and parabola y2 = x, then the value of $8|{m_1}{m_2}|$ is equal to :
Let $x = 2t$, $y = {{{t^2}} \over 3}$ be a conic. Let S be the focus and B be the point on the axis of the conic such that $SA \bot BA$, where A is any point on the conic. If k is the ordinate of the centroid of the $\Delta$SAB, then $\mathop {\lim }\limits_{t \to 1} k$ is equal to :
A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with :
Let x2 + y2 + Ax + By + C = 0 be a circle passing through (0, 6) and touching the parabola y = x2 at (2, 4). Then A + C is equal to ___________.
Two tangent lines $l_{1}$ and $l_{2}$ are drawn from the point $(2,0)$ to the parabola $2 \mathrm{y}^{2}=-x$. If the lines $l_{1}$ and $l_{2}$ are also tangent to the circle $(x-5)^{2}+y^{2}=r$, then 17r is equal to ___________.
Explanation:
Given : ${y^2} = {{ - x} \over 2}$
$\eqalign{ & T \equiv y = mx - {1 \over {8m}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow (2,0) \cr} $
$ \Rightarrow {m^2} = {1 \over {16}} \Rightarrow m = \, \pm \,{1 \over 4}$
Tangents are $y = {1 \over 4}x - {1 \over 2},\,y = {{ - x} \over 4} + {1 \over 2}$
$4y = x - 2$ and $4y + x = 2$
If these are also tangent to circle then ${d_c} = r$
$ \Rightarrow \left| {{{5 - 2} \over {\sqrt {17} }}} \right| = \sqrt r \Rightarrow r = {\left( {{3 \over {\sqrt {17} }}} \right)^2}$
$ \Rightarrow 17r = 17\,.\,{9 \over {17}} = 9$
The sum of diameters of the circles that touch (i) the parabola $75 x^{2}=64(5 y-3)$ at the point $\left(\frac{8}{5}, \frac{6}{5}\right)$ and (ii) the $y$-axis, is equal to ______________.
Explanation:
Equation of tangent to the parabola at $P\left(\frac{8}{5}, \frac{6}{5}\right)$
$ \begin{aligned} &75 x \cdot \frac{8}{5}=160\left(y+\frac{6}{5}\right)-192 \\\\ &\Rightarrow 120 x=160 y \\\\ &\Rightarrow 3 x=4 y \end{aligned} $
Equation of circle touching the given parabola at $\mathrm{P}$ can be taken as
$\left(x-\frac{8}{5}\right)^{2}+\left(y-\frac{6}{5}\right)^{2}+\lambda(3 x-4 y)=0$
If this circle touches $y$-axis then
$ \begin{aligned} &\frac{64}{25}+\left(y-\frac{6}{5}\right)^{2}+\lambda(-4 y)=0 \\\\ &\Rightarrow y^{2}-2 y\left(2 \lambda+\frac{6}{5}\right)+4=0 \\\\ &\Rightarrow D=0 \\\\ &\Rightarrow\left(2 \lambda+\frac{6}{6}\right)^{2}=4 \\\\ &\Rightarrow \lambda=\frac{2}{5} \text { or }-\frac{8}{5} \end{aligned} $
Radius $=1$ or 4
Sum of diameter $=10$
Let PQ be a focal chord of length 6.25 units of the parabola y2 = 4x. If O is the vertex of the parabola, then 10 times the area (in sq. units) of $\Delta$POQ is equal to ___________.
Explanation:
Given parabola ${y^2} = 4x$
$\therefore$ a = 1
Here, P, S, Q points are collinear.
$\therefore$ Slope of PS = Slope of QS
$ \Rightarrow {{2{t_1} - 0} \over {t_1^2 - 1}} = {{0 - 2{t_2}} \over {1 - t_2^2}}$
$ \Rightarrow {{2{t_1}} \over {t_1^2 - 1}} = {{2{t_2}} \over {t_2^2 - 1}}$
$ \Rightarrow {t_1}(t_2^2 - 1) = {t_2}(t_1^2 - 1)$
$ \Rightarrow t_2^2{t_1} - {t_1} = t_1^2{t_2} - {t_2}$
$ \Rightarrow t_2^2{t_1} - t_1^2{t_2} - {t_1} + {t_2} = 0$
$ \Rightarrow {t_1}{t_2}({t_2} - {t_1}) + ({t_2} - {t_1}) = 0$
$ \Rightarrow ({t_2} - {t_1})({t_1}{t_2} + 1) = 0$
As ${t_2} - {t_1} \ne 0$
$\therefore$ ${t_1}{t_2} + 1 = 0$
${t_1}{t_2} = - 1$
Now, lenght of PQ
$ = \sqrt {{{\left( {t_1^2 - t_2^2} \right)}^2} + {{\left( {2{t_1} - 2{t_2}} \right)}^2}} $
$ = \sqrt {{{\left( {{t_1} + {t_2}} \right)}^2}{{\left( {{t_1} - {t_2}} \right)}^2} + 4{{\left( {{t_1} - {t_2}} \right)}^2}} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {{{\left( {{t_1} + {t_2}} \right)}^2} + 4} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2{t_1}{t_2} + 4} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2( - 1) + 4} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 - 2( - 1)} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 - 2{t_1}{t_2}} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {{{\left( {{t_1} - {t_2}} \right)}^2}} $
$ = {\left( {{t_1} - {t_2}} \right)^2}$
Given, length of $PQ = {\left( {{t_1} - {t_2}} \right)^2} = 6.25$
$ \Rightarrow {t_1} - {t_2} = 2.5$
Now, Area of $\Delta OPQ$
$ = \left| {{1 \over 2}\left| {\matrix{ {t_1^2} & {2{t_1}} & 1 \cr {t_2^2} & {2{t_2}} & 1 \cr 0 & 0 & 1 \cr } } \right|} \right|$
$ = \left| {{1 \over 2}\left( {2{t_2}t_1^2 - 2{t_1}t_2^2} \right)} \right|$
$ = \left| {{1 \over 2} \times 2{t_1}{t_2}\left( {{t_1} - {t_2}} \right)} \right|$
$ = \left| {{t_1}{t_2} \times \left( {{t_1} - {t_2}} \right)} \right|$
$ = \left| { - 1 \times 2.5} \right|$
$ = 2.5$
$\therefore$ 10$\Delta$OPQ =
$ = 10 \times {{25} \over {10}} = 25$
A circle of radius 2 unit passes through the vertex and the focus of the parabola y2 = 2x and touches the parabola $y = {\left( {x - {1 \over 4}} \right)^2} + \alpha $, where $\alpha$ > 0. Then (4$\alpha$ $-$ 8)2 is equal to ______________.
Explanation:
Let the equation of circle be
$x\left( {x - {1 \over 2}} \right) + {y^2} + \lambda y = 0$
$ \Rightarrow {x^2} + {y^2} - {1 \over 2}x + \lambda y = 0$
Radius $ = \sqrt {{1 \over {16}} + {{{\lambda ^2}} \over 4}} = 2$
$ \Rightarrow {\lambda ^2} = {{63} \over 4}$
$ \Rightarrow {\left( {x - {1 \over 4}} \right)^2} + {\left( {y + {\lambda \over 2}} \right)^2} = 4$
$\because$ This circle and parabola $y - \alpha = {\left( {x - {1 \over 4}} \right)^2}$ touch each other, so
$\alpha = - {\lambda \over 2} + 2$
$ \Rightarrow \alpha - 2 = - {\lambda \over 2}$
$ \Rightarrow {(\alpha - 2)^2} = {{{\lambda ^2}} \over 4} = {{63} \over {16}}$
$ \Rightarrow {(4\alpha - 8)^2} = 63$
Let the common tangents to the curves $4({x^2} + {y^2}) = 9$ and ${y^2} = 4x$ intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and l respectively denote the eccentricity and the length of the latus rectum of this ellipse, then ${l \over {{e^2}}}$ is equal to ______________.
Explanation:
Let y = mx + c is the common tangent
So $c = {1 \over m} = \pm \,{3 \over 2}\sqrt {1 + {m^2}} \Rightarrow {m^2} = {1 \over 3}$
So equation of common tangents will be $y = \pm \,{1 \over {\sqrt 3 }}x \pm \,\sqrt 3 $, which intersects at Q($-$3, 0)
Major axis and minor axis of ellipse are 12 and 6.
So eccentricity
${e^2} = 1 - {1 \over 4} = {3 \over 4}$ and length of latus rectum $ = {{2{b^2}} \over a} = 3$
Hence, ${l \over {{e^2}}} = {3 \over {3/4}} = 4$
Let P1 be a parabola with vertex (3, 2) and focus (4, 4) and P2 be its mirror image with respect to the line x + 2y = 6. Then the directrix of P2 is x + 2y = ____________.
Explanation:
Focus = (4, 4) and vertex = (3, 2)
$\therefore$ Point of intersection of directrix with axis of parabola = A = (2, 0)
Image of A(2, 0) with respect to line x + 2y = 6 is B(x2, y2)
$\therefore$ ${{{x_2} - 2} \over 1} = {{{y_2} - 0} \over 2} = {{ - 2(2 + 0 - 6)} \over 5}$
$\therefore$ $B({x_2},\,{y_2}) = \left( {{{18} \over 5},{{16} \over 5}} \right)$.
Point B is point of intersection of direction with axes of parabola P2.
$\therefore$ $x + 2y = \lambda $ must have point $\left( {{{18} \over 5},{{16} \over 5}} \right)$
$\therefore$ $x + 2y = 10$
parabola y2 = 16(x $-$ 3) are at right angles, then the locus of point P is :
Explanation:
P(2, $-$4) $\Rightarrow$ $-$4 = 2m + ${2 \over m}$
$\Rightarrow$ m + ${1 \over m}$ = $-$2 $\Rightarrow$ m = $-$1
$\therefore$ tangent is y = $-$x $-$2
$\Rightarrow$ x + y + 2 = 0 ...... (1)
(1) is also tangent to x2 + y2 = a
So, ${2 \over {\sqrt 2 }} = \sqrt a \Rightarrow \sqrt a = \sqrt 2 $
$\Rightarrow$ a = 2
Explanation:
Normal at point P
$tx + y = 3t + {3 \over 2}{t^3}$
Passes through $\left( {3,{3 \over 2}} \right)$
$ \Rightarrow 3t + {3 \over 2} = 3t + {3 \over 2}{t^3}$
$ \Rightarrow {t^3} = 1 \Rightarrow t = 1$
$P \equiv \left( {{3 \over 2},3} \right) = (\alpha ,\beta )$
$2(\alpha + \beta ) = 2\left( {{3 \over 2} + 3} \right) = 9$
Explanation:
focus : ($-$16, 0)
y = mx + c is focal chord
$\Rightarrow$ c = 16 m ...........(1)
y = mx + c is tangent to (x + 10)2 + y2 = 4
$\Rightarrow$ y = m(x + 10) $\pm$ 2$\sqrt {1 + {m^2}} $
$\Rightarrow$ c = 10m $\pm$ 2$\sqrt {1 + {m^2}} $
$\Rightarrow$ 16m = 10m $\pm$ 2$\sqrt {1 + {m^2}} $
$\Rightarrow$ 6m = 2$\sqrt {1 + {m^2}} $ (m > 0)
$\Rightarrow$ 9m2 = 1 + m2
$\Rightarrow$ m = ${1 \over {2\sqrt 2 }}$ & c = ${8 \over {\sqrt 2 }}$
$4\sqrt 2 (m + c) = 4\sqrt 2 \left( {{{17} \over {2\sqrt 2 }}} \right)$ = 34
Explanation:
Parabola : y2 = 4x
Let tangent y = mx + ${a \over m}$
y = mx + ${1 \over m}$
m2x $-$ my + 1 = 0
the above line is also tangent to circle
(x $-$ 3)2 + y2 = 9
$\therefore$ $ \bot $ from (3, 0) = 3
$\left| {{{3{m^2} - 0 + 1} \over {\sqrt {{m^2} + {m^4}} }}} \right| = 3$
(3m2 + 1)2 = 9(m2 + m4)
$6{m^2} + 1 + 9{m^4} = 9{m^2} + 9{m^4}$
$3{m^2} = 1$
$m = \pm {1 \over {\sqrt 3 }}$
$ \therefore $ tangent is
$y = {1 \over {\sqrt 3 }}x + \sqrt 3 $
(it will be used)
or
$y = - {1 \over {\sqrt 3 }}x - \sqrt 3 $
(rejected)
$m = {1 \over {\sqrt 3 }}$
For parabola
$\left( {{a \over {{m^2}}},{{2a} \over m}} \right) \equiv (3,2\sqrt 3 )$ = (c, d)
for circle $y = {1 \over {\sqrt 3 }}x + \sqrt 3 $
&
${(x - 3)^2} + {y^2} = 9$
Solving,
${(x - 3)^2} + {\left( {{1 \over {\sqrt 3 }}x + \sqrt 3 } \right)^2} = 9$
${x^2} + 9 - 6x + {1 \over 3}{x^2} + 3 + 2x = 9$
${4 \over 3}{x^2} - 4x + 3 = 0$
$4{x^2} - 12x + 9 = 0$
$4{x^2} - 6x - 6x + 9 = 0$
$2x(2x - 3) - 3(2x - 3) = 0$
$(2x - 3)(2x - 3) = 0$
$x = {3 \over 2}$
$ \therefore $ $y = {1 \over {\sqrt 3 }}\left( {{3 \over 2}} \right) + \sqrt 3 $
$y = {{\sqrt 3 } \over 2} + \sqrt 3 $
$y = {{3\sqrt 3 } \over 2}$
$(a,b) \equiv \left( {{3 \over 2},{{3\sqrt 3 } \over 2}} \right)$
$2(a + c) = 2\left( {{3 \over 2} + 3} \right)$
$ = 2\left( {{3 \over 2} + {6 \over 2}} \right) = 9$
y = x2 at the point (2, 4) is :
and L2 be a tangent to the parabola y2 = 8(x + 2)
such that L1 and L2 intersect at right angles. Then L1 and L2 meet on the straight line :
y2 = 4x and x2 = 4y also touches the circle, x2 + y2 = c2,
then c is equal to :