Let $(h, k)$ lie on the circle $\mathrm{C}: x^2+y^2=4$ and the point $(2 h+1,3 k+2)$ lie on an ellipse with eccentricity $e$. Then the value of $\frac{5}{e^2}$ is equal to $\_\_\_\_$ .
Explanation:
$(\mathrm{h}, \mathrm{k})$ lie on circle $x^2+y^2=1$.
so, $h^2+k^2=1--$ (1)
$(2 h+1), 3 k+2)$ lies on ellipse This means $x=2 h+1$ and $y=3 k+2$ lies on ellipse
Substitute $\mathrm{h}=\frac{x-1}{2}$ and $\mathrm{k}=\frac{y-2}{3}$ in equation (1)
$ \begin{aligned} & \left(\frac{x-1}{2}\right)^2+\left(\frac{y-2}{3}\right)^2=1 \\ & \frac{(x-1)^2}{4}+\frac{(y-2)^2}{9}=1 \end{aligned} $
This is shifted ellipse of the form $\frac{\left(x-x_1\right)^2}{-2}+\frac{\left(y-y_1\right)^2}{12}=1 \mathrm{a}=2, \mathrm{~b}=3$
eccentricity e $=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}$
$e^2=\frac{5}{9} \Longrightarrow \frac{5}{e^2}=9$ Ans
Let $\mathrm{E}_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ be an ellipse. Ellipses $\mathrm{E}_{\mathrm{i}}$ 's are constructed such that their centres and eccentricities are same as that of $\mathrm{E}_1$, and the length of minor axis of $\mathrm{E}_{\mathrm{i}}$ is the length of major axis of $E_{i+1}(i \geq 1)$. If $A_i$ is the area of the ellipse $E_i$, then $\frac{5}{\pi}\left(\sum\limits_{i=1}^{\infty} A_i\right)$, is equal to _______.
Explanation:
$\begin{aligned} & E_1=\frac{x^2}{9}+\frac{y^2}{4} \Rightarrow e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3} \\ & E_2: \frac{x^2}{a^2}+\frac{y^2}{4}=1 \\ & e=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{a^2}{4}} \Rightarrow \frac{5}{9}=1-\frac{a^2}{4} \\ & a^2=\frac{16}{9} \\ & E_2: \frac{x^2}{16}+\frac{y^2}{4}=1 \\ & E_3: \frac{x^2}{\frac{16}{9}}+\frac{y^2}{b^2}=1 \\ & e=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{b^2}{16}} \Rightarrow b^2=\frac{64}{81} \end{aligned}$
$\begin{aligned} & \mathrm{E}_3=\frac{\mathrm{x}^2}{\frac{16}{9}}+\frac{\mathrm{y}^2}{\frac{64}{81}}=1 \\ & \mathrm{~A}_1=\pi \times 3 \times 2 \Rightarrow 6 \pi \\ & \mathrm{~A}_2=\pi \times \frac{4}{3} \times 2=\frac{8 \pi}{3} \\ & \mathrm{~A}_3=\pi \times \frac{4}{3} \times \frac{8}{9}=\frac{32 \pi}{81} \\ & \sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}}=6 \pi+\frac{8 \pi}{3}+\frac{32 \pi}{81}+\ldots \infty \Rightarrow \frac{6 \pi}{1-\frac{4}{9}} \Rightarrow \frac{54 \pi}{5} \\ & \therefore \frac{5}{\pi} \sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}} \Rightarrow \frac{5}{\pi} \times \frac{54 \pi}{5}=54 \end{aligned}$
Explanation:
$ \begin{aligned} & \frac{2 b^2}{a}=\frac{1}{2}, \quad \tan 30^{\circ}=\frac{b}{a e} \\\\ & b^2=\frac{a}{4}, \frac{1}{3}=\frac{b^2}{a^2-b^2} \Rightarrow a^2-b^2=3 b^2 \Rightarrow b^2=\frac{a^2}{4} \\\\ & \Rightarrow \quad a=1, b^2=\frac{1}{4} \Rightarrow b=\frac{1}{2} \\\\ & \Rightarrow \quad(2 a+2 b)^2=9 \end{aligned} $
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 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)
$=2525\left(\frac{4}{25}+1\right)$
$=2929$
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$
Explanation:
$\therefore$ ${M_1}{M_2} = {1 \over 2}QQ'$
Maximum value of QQ' is AA'
Hence, maximum value of ${M_1}{M_2} = {1 \over 2}AA' = 4$
Explanation:
We will consider three cases
Case 1 : Circle passes through origin, that is, p = 0 the equation of circle becomes
x2 + y2 + 2x + 4y = 0
x = 0 $\Rightarrow$ y2 + 4y = 0 $\Rightarrow$ y(y + 4) = 0
y = 0, $-$4
y = 0 $\Rightarrow$ x2 + 2x = 0 $\Rightarrow$ x(x + 2) = 0 $\Rightarrow$ x = 0, $-$ 2
Case 2 : Circle touches y axis then circle will intersect x axis at two distinct points
Put y = 0 in equation of circle, we get
x2 + 2x $-$ p = 0
Now from g2 $-$ c > 0 and f2 $-$ c = 0
$\Rightarrow$ 12 $-$ ($-$p) > 0 and 22 $-$ ($-$p) = 0
1 + p > 0 and 4 + p = 0
p > $-$1 and p = $-$ 4
which is a contradiction. Also, for p = $-$4 we get
x2 $-$ 2x + 4 = 0
$x = {{2 \pm \sqrt {4 - 4 \times 4} } \over 2} = {{2 \pm 2\sqrt 3 i} \over 2} = 1 \pm \sqrt 3 i$
They are imaginary roots.
Therefore, Case 2 is not possible.
Case 3 : Circle touches x axis. Then the circle will intersect y axis at two distinct points.
Substituting x = 0, we get
y2 + 4y $-$ p = 0
Now, from g2 $-$ C = 0 and f2 $-$ C > 0, we have
12 $-$ ($-$p) = 0 and 22 = ($-$p) > 0
p = $-$1 and 4 + p > 0 $\Rightarrow$ p > $-$4
Therefore, the equation of circle becomes
y2 + 4y + 1 = 0
$ \Rightarrow y = {{ - 4 \pm \sqrt {16 - 4} } \over 2} = {{ - 4 \pm \sqrt {12} } \over 2}$
$y = {{ - 4 \pm \sqrt {4 \times 3} } \over 2}$
$ = {{ - 4 \pm 2\sqrt 3 } \over 2} = - 2 \pm \sqrt 3 $ real values
Thus, Case 3 is possible.
Thus, for p = 0 and p = $-$1 the circle and coordinates have exactly three points in common. Hence, the correct answer is 2.
Explanation:
A vertical line passing through $(h, 0)$ is $x=h$ and this vertical line intersect the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at P and Q.
Put $x=h$ in the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$
$\begin{aligned} & \Rightarrow \frac{h^2}{4}+\frac{y^2}{3}=1 \\ & \Rightarrow y= \pm \sqrt{3-\frac{3 h^2}{4}} \end{aligned}$
So $\mathrm{P}=\left(h, \sqrt{3-\frac{3 h^2}{4}}\right)$ and $\mathrm{Q}=\left(h,-\sqrt{3-\frac{3 h^2}{4}}\right)$
Equation of tangent of ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at P is
$\frac{h x}{4}+\frac{y}{3} \sqrt{3-\frac{3 h^2}{4}}=1 \quad \text{... (i)}$
Equation of tangent of ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at Q is
$\frac{h x}{4}-\frac{y}{3} \sqrt{3-\frac{3 h^2}{4}}=1 \quad \text{... (ii)}$
Given, $R$ is the point of intersection of tangents at $P$ and at $Q$.
On solving the equations (i) and (ii)
$\Rightarrow \quad \mathrm{R}=\left(\frac{8}{h}, 0\right)$
Given, $\Delta(h)=$ Area of $\Delta \mathrm{PQR}$
$\begin{aligned} \Rightarrow \quad \Delta(h) & =\frac{1}{2}\left(\frac{8}{h}-h\right) \cdot 2 \sqrt{3-\frac{3 h^2}{4}} \\ \Rightarrow \quad \Delta(h) & =\frac{\sqrt{3}}{2}\left(\frac{8}{h}-h\right) \sqrt{4-h^2} \\ \Rightarrow \quad \frac{d \Delta(h)}{d h}= & \frac{\sqrt{3}}{2}\left(\frac{-8}{h^2}-1\right) \sqrt{4-h^2} +\frac{\sqrt{3}}{2}\left(\frac{8}{h}-h\right) \cdot \frac{(-2 h)}{2 \sqrt{4-h^2}} \end{aligned}$
$\begin{aligned} \Rightarrow \quad \frac{d \Delta(h)}{d h}= & \frac{-\sqrt{3}}{2 \sqrt{4-h^2}} \\ & {\left[\left(\frac{8}{h^2}+1\right)\left(4-h^2\right)+\left(\frac{8}{h}-h\right) h\right] } \end{aligned}$
$\begin{aligned} & \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{-\sqrt{3}}{2 \sqrt{4-h^2}}\left[\frac{32}{h^2}-8+4-\hbar^2+8-h^2\right] \\ & \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{-\sqrt{3}}{2 \sqrt{4-h^2}}\left[-2 h^2+\frac{32}{h^2}+4\right] \\ & \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{\sqrt{3}}{h^2 \sqrt{4-h^2}}\left[h^4-2 h^2-16\right] \\ & \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{\sqrt{3}}{h^2 \sqrt{4-h^2}}\left[\left(h^2-1\right)^2-17\right] \end{aligned}$
For, $\frac{1}{2} \leq h \leq 1, \quad \frac{d \Delta(h)}{d h}<0$
$\therefore \quad \Delta(h)$ is a decreasing function for $x \in\left[\frac{1}{2}, 1\right]$
$\Rightarrow \Delta(1) \leq \Delta(h) \leq \Delta\left(\frac{1}{2}\right)$
Given $\quad \Delta_1=\max _{\frac{1}{2} \leq h \leq 1} \Delta(h)$ and
$\begin{aligned} \Delta_2 & =\min _{\frac{1}{2} \leq h \leq 1} \Delta(h) \\ \Rightarrow \quad \Delta_1 & =\Delta\left(\frac{1}{2}\right) \text { and } \Delta_2=\Delta(1) \\ \Rightarrow \quad \Delta_1 & =\frac{\sqrt{3}}{2}\left(16-\frac{1}{2}\right) \sqrt{4-\frac{1}{4}} \text { and } \\ \Delta_2 & =\frac{\sqrt{3}}{2}(8-1) \sqrt{4-1} \\ \Rightarrow \quad \Delta_1 & =\frac{\sqrt{3}}{2} \times \frac{31}{2} \times \frac{\sqrt{5} \cdot \sqrt{3}}{2} \text { and } \Delta_2=\frac{21}{2} \\ \Rightarrow \quad \frac{8 \Delta_1}{\sqrt{5}} & =93 \text { and } 8 \Delta_2=84 \end{aligned}$
$\Rightarrow \frac{8 \Delta_1}{\sqrt{5}}-8 \Delta_2=9$
Hints:
(i) If $f(x)$ is a decreasing function in $x \in[a, b]$, then $f(b) \leq f(x) \leq f(a)$
(ii) If $\left(x_1, y_1\right)$ is a point on the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then equation of tangent at $\left(x_1, y_1\right)$ is $\frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1$.