Ellipse
The line $x=8$ is the directrix of the ellipse $\mathrm{E}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with the corresponding focus $(2,0)$. If the tangent to $\mathrm{E}$ at the point $\mathrm{P}$ in the first quadrant passes through the point $(0,4\sqrt3)$ and intersects the $x$-axis at $\mathrm{Q}$, then $(3\mathrm{PQ})^{2}$ is equal to ____________.
Explanation:
$\begin{aligned} & \mathrm{P}(2 \sqrt{3}, \sqrt{3}) \\\\ & \mathrm{Q}\left(\frac{8}{\sqrt{3}}, 0\right) \\\\ & (3 \mathrm{PQ})^2=39\end{aligned}$
Let C be the largest circle centred at (2, 0) and inscribed in the ellipse ${{{x^2}} \over {36}} + {{{y^2}} \over {16}} = 1$. If (1, $\alpha$) lies on C, then 10 $\alpha^2$ is equal to ____________
Explanation:
$r^{2}=(x-2)^{2}+y^{2}$
Solving simultaneously
$-5 x^{2}+36 x+\left(9 r^{2}-180\right)=0$
$D=0$
$r^{2}=\frac{128}{10}$
Distance between $(1, \alpha)$ and $(2,0)$ should be $r$
$ \begin{aligned} & 1+\alpha^{2}=\frac{128}{10} \\\\ & \alpha^{2}=\frac{118}{10} \\\\ &=118.00 \end{aligned} $
Let a tangent to the curve $9{x^2} + 16{y^2} = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is ________
Explanation:
Given curve,
$9{x^2} + 16{y^2} = 144$
$ \Rightarrow {{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$
$ \Rightarrow {{{x^2}} \over {{4^2}}} + {{{y^2}} \over {{3^2}}} = 1$
$\therefore$ a = 4 and b = 3
So, general point on the ellipse is $ = (4\cos \theta ,3\sin \theta )$
We know,
Equation of tangent to a given ellipse at its point $(a\cos\theta ,b\sin \theta )$ is
${{x\cos \theta } \over a} + {{y\sin \theta } \over b} = 1$
$\therefore$ Here equation of tangent at point $(4\cos \theta ,3\sin \theta )$ is
${{x\cos \theta } \over 4} + {{y\sin \theta } \over 3} = 1$
When this tangent cut's x axis then y = 0.
$\therefore$ ${{x\cos \theta } \over 4} + 0 = 1$
$ \Rightarrow x = 4\sec \theta $
$\therefore$ Point of intersection at x axis is $A(4\sec \theta ,0)$.
When this tangent cut's y axis then x = 0.
$\therefore$ $0 + {{y\sin \theta } \over 3} = 1$
$ \Rightarrow y = 3\cos ec\theta $
$\therefore$ Point of intersection at y axis is $B(0,3\cos ec\theta )$
$\therefore$ Length of AB
$ = \sqrt {{{(4\sec \theta - 0)}^2} + {{(0 - 3\cos ec\theta )}^2}} $
$ = \sqrt {16{{\sec }^2}\theta + 9\cos e{c^2}\theta } $
$ = \sqrt {16(1 + {{\tan }^2}\theta ) + 9(1 + {{\cot }^2}\theta )} $
$ = \sqrt {25 + 16{{\tan }^2}\theta + 9{{\cot }^2}\theta } $
We know, $AM \ge GM$
$\therefore$ ${{16{{\tan }^2}\theta + 9{{\cot }^2}\theta } \over 2} \ge \sqrt {(16{{\tan }^2}\theta )(9{{\cot }^2}\theta )} $
$ \Rightarrow 16{\tan ^2}\theta + 9{\cot ^2}\theta \ge 2(4\tan \theta )(3\cot \theta )$
$ \Rightarrow 16{\tan ^2}\theta + 9{\cot ^2}\theta \ge 2 \times 4 \times 3$
$ \Rightarrow 16{\tan ^2}\theta + 9{\cot ^2}\theta \ge 24$
$\therefore$ $AB = \sqrt {25 + 16{{\tan }^2}\theta + 9{{\cot }^2}\theta } $
$ \ge \sqrt {25 + 24} $
$ \ge \sqrt {49} $
$ \ge 7$
$\therefore$ Minimum length of $AB = 7$.
Let a line L pass through the point of intersection of the lines $b x+10 y-8=0$ and $2 x-3 y=0, \mathrm{~b} \in \mathbf{R}-\left\{\frac{4}{3}\right\}$. If the line $\mathrm{L}$ also passes through the point $(1,1)$ and touches the circle $17\left(x^{2}+y^{2}\right)=16$, then the eccentricity of the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{\mathrm{~b}^{2}}=1$ is :
The acute angle between the pair of tangents drawn to the ellipse $2 x^{2}+3 y^{2}=5$ from the point $(1,3)$ is :
If the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ meets the line $\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$ on the $x$-axis and the line $\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$ on the $y$-axis, then the eccentricity of the ellipse is :
Let the eccentricity of the ellipse ${x^2} + {a^2}{y^2} = 25{a^2}$ be b times the eccentricity of the hyperbola ${x^2} - {a^2}{y^2} = 5$, where a is the minimum distance between the curves y = ex and y = logex. Then ${a^2} + {1 \over {{b^2}}}$ is equal to :
Let the eccentricity of an ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$, $a > b$, be ${1 \over 4}$. If this ellipse passes through the point $\left( { - 4\sqrt {{2 \over 5}} ,3} \right)$, then ${a^2} + {b^2}$ is equal to :
If m is the slope of a common tangent to the curves ${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$ and ${x^2} + {y^2} = 12$, then $12{m^2}$ is equal to :
The locus of the mid point of the line segment joining the point (4, 3) and the points on the ellipse ${x^2} + 2{y^2} = 4$ is an ellipse with eccentricity :
The line y = x + 1 meets the ellipse ${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$ at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)2 is equal to :
Let the maximum area of the triangle that can be inscribed in the ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 4} = 1,\,a > 2$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $6\sqrt 3 $. Then the eccentricity of the ellipse is :
Let the tangents at the points $\mathrm{P}$ and $\mathrm{Q}$ on the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$ meet at the point $R(\sqrt{2}, 2 \sqrt{2}-2)$. If $\mathrm{S}$ is the focus of the ellipse on its negative major axis, then $\mathrm{SP}^{2}+\mathrm{SQ}^{2}$ is equal to ___________.
Explanation:
$E \equiv {{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$
$\eqalign{ & T \equiv y = mx\, \pm \,\sqrt {2{m^2} + 4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \left( {\sqrt 2 ,2\sqrt 2 - 2} \right) \cr} $
$ \Rightarrow \left( {2\sqrt 2 - 2 - m\sqrt 2 } \right) = \pm \,\sqrt {2{m^2} + 4} $
$ \Rightarrow 2{m^2} - 2m\sqrt 2 \left( {2\sqrt {2 - 2} } \right) + 4(3 - 2\sqrt 2 ) = 2{m^2} + 4$
$ \Rightarrow - 2\sqrt 2 m(2\sqrt 2 - 2) = 4 - 12 + 8\sqrt 2 $
$ \Rightarrow - 4\sqrt 2 m(\sqrt 2 - 1) = 8(\sqrt 2 - 1)$
$ \Rightarrow m = - \sqrt 2 $ and $m \to \infty $
$\therefore$ Tangents are $x = \sqrt 2 $ and $y = - \sqrt 2 x + \sqrt 8 $
$\therefore$ $P(\sqrt 2 ,0)$ and $Q(1,\sqrt 2 )$
and $S = (0, - \sqrt 2 )$
$\therefore$ ${(PS)^2} + {(QS)^2} = 4 + 9 = 13$
If the length of the latus rectum of the ellipse $x^{2}+4 y^{2}+2 x+8 y-\lambda=0$ is 4 , and $l$ is the length of its major axis, then $\lambda+l$ is equal to ____________.
Explanation:
Equation of ellipse is : ${x^2} + 4{y^2} + 2x + 8y - \lambda = 0$
${(x + 1)^2} + 4{(y + 1)^2} = \lambda + 5$
${{{{(x + 1)}^2}} \over {\lambda + 5}} + {{{{(y + 1)}^2}} \over {\left( {{{\lambda + 5} \over 4}} \right)}} = 1$
Length of latus rectum $ = {{2\,.\,\left( {{{\lambda + 5} \over 4}} \right)} \over {\sqrt {\lambda + 5} }} = 4$.
$\therefore$ $\lambda = 59$.
Length of major axis $ = 2\,.\,\sqrt {\lambda + 5} = 16 = l$
$\therefore$ $\lambda + l = 75$.
If two tangents drawn from a point ($\alpha$, $\beta$) lying on the ellipse 25x2 + 4y2 = 1 to the parabola y2 = 4x are such that the slope of one tangent is four times the other, then the value of (10$\alpha$ + 5)2 + (16$\beta$2 + 50)2 equals ___________.
Explanation:
Tangent to the parabola, $y=m x+\frac{1}{m}$ passes through $(\alpha, \beta)$. So, $\alpha m^{2}-\beta m+1=0$ has roots $m_{1}$ and $4 m_{1}$,
$ m_{1}+4 m_{1}=\frac{\beta}{\alpha} \text { and } m_{1} \cdot 4 m_{1}=\frac{1}{\alpha} $
Gives that $4 \beta^{2}=25 \alpha \quad\quad...(ii)$
from (i) and (ii)
$25\left(\alpha^{2}+\alpha\right)=1\quad\quad...(iii)$
Now, $(10 \alpha+5)^{2}+\left(16 \beta^{2}+50\right)^{2}$
$=25(2 \alpha+1)^{2}+2500(2 \alpha+1)^{2}$
$=2525\left(4 \alpha^{2}+4 \alpha+1\right)$ from equation (iii)
$=2525\left(\frac{4}{25}+1\right)$
$=2929$
circle x2 + y2 = 4b, b > 4 lie on the curve y2 = 3x2, then b is equal to :
Explanation:
${{x\cos \theta } \over b} + {{y\sin \theta } \over {2a}} = 1$
So, area $(\Delta OAB) = {1 \over 2} \times {b \over {\cos \theta }} \times {{2a} \over {\sin \theta }}$
$ = {{2ab} \over {\sin 2\theta }} \ge 2ab$
$\Rightarrow$ k = 2
Explanation:
and A(5, $-$4)
Hence, a = 2 & ae = 1
$\Rightarrow$ e = ${1 \over 2}$
$\Rightarrow$ b2 = 3
So, $E:{{{{(x - 3)}^2}} \over 4} + {{{{(y + 4)}^2}} \over 3} = 1$
Intersecting with given tangent.
${{{x^2} - 6x + 9} \over 4} + {{{m^2}{x^2}} \over 3} = 1$
Now, D = 0 (as it is tngent)
So, 5m2 = 3.
4x2 + 9y2 = 36 and (2x)2 + (2y)2 = 31. Then the
square of the slope of the line L is __________.
Explanation:
$y = mx + \sqrt {9{m^2} + 4} $
and equation of tangent to the curve ${x^2} + {y^2} = {{31} \over 4}$ is
$y = mx + \sqrt {{{31} \over 4}{{(1 + m)}^2}} $
for common tangent $9{m^2} + 4 = {{31} \over 4} + {{31} \over 4}{m^2}$
$ \Rightarrow {5 \over 4}{m^2} = {{15} \over 4}$
$ \Rightarrow {m^2} = 3$
${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$
from any of its foci?
are $\left( {\sqrt 7 ,0} \right)$ and $\left( { - \sqrt 7 ,0} \right)$ respectively and
P is any point on the conic, 9x2 + 16y2 = 144, then PA + PB is equal to :
$\phi \left( t \right) = {5 \over {12}} + t - {t^2}$, then a2 + b2 is equal to :
${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1$ for some $a$ $ \in $ R, then the distance between the foci of the ellipse is :
