Question

Consider an ellipse with major & minor axis having length 4 & $2\sqrt{3}$ respectively. Let P be any variable point on it. S & S'be the foci and C be the centre of ellipse Match List-I with List-II and select the correct answer using the code given below the list. List-I List-II (P) Maximum value of PS.PS' is (Q) Minimum value of $(PS)^2+(PS')^2$ is (R) Minimum length of portion of tangent at P intercepted between principal axes is (S) If foot of normal from P on major axis is P' and normal at P cuts transverse axis at N, then $\frac{CN}{CP'}$ is

• A. 3
• B. 1
• C. 2
• D. 4
• E. 8
• F. $2+\sqrt{3}$
• G. 4
• H. $2+\sqrt{3}$
• I. 1/4

Answer 1

Answer: (G), (I)

P-3, Q-1, R-2, S-4

Answer 2

The ellipse has a major axis of length $2a = 4$, so $a=2$. The minor axis has length $2b = 2\sqrt{3}$, so $b=\sqrt{3}$. The eccentricity $e$ is given by $b^2 = a^2(1-e^2)$. $(\sqrt{3})^2 = 2^2(1-e^2) \implies 3 = 4(1-e^2) \implies 1-e^2 = \frac{3}{4} \implies e^2 = \frac{1}{4} \implies e = \frac{1}{2}$. The foci are $S=(ae, 0)$ and $S'=(-ae, 0)$. $ae = 2 \times \frac{1}{2} = 1$. So $S=(1,0)$ and $S'=(-1,0)$. The center $C$ is $(0,0)$. Let $P(x,y)$ be any point on the ellipse. The distances from P to the foci are $PS = a+ex$ and $PS' = a-ex$. For $P(a\cos\theta, b\sin\theta) = (2\cos\theta, \sqrt{3}\sin\theta)$, $x = 2\cos\theta$. $PS = 2 + \frac{1}{2}(2\cos\theta) = 2+\cos\theta$. $PS' = 2 - \frac{1}{2}(2\cos\theta) = 2-\cos\theta$.

(P) Maximum value of PS.PS' $PS \cdot PS' = (2+\cos\theta)(2-\cos\theta) = 4 - \cos^2\theta$. This product is maximum when $\cos^2\theta$ is minimum, i.e., $\cos^2\theta = 0$. Maximum value $= 4 - 0 = 4$. This matches with option (3).

(Q) Minimum value of $(PS)^2+(PS')^2$ $(PS)^2 + (PS')^2 = (2+\cos\theta)^2 + (2-\cos\theta)^2$ $= (4 + 4\cos\theta + \cos^2\theta) + (4 - 4\cos\theta + \cos^2\theta)$ $= 8 + 2\cos^2\theta$. This sum is minimum when $\cos^2\theta$ is minimum, i.e., $\cos^2\theta = 0$. Minimum value $= 8 + 2(0) = 8$. This matches with option (1).

(R) Minimum length of portion of tangent at P intercepted between principal axes The equation of the tangent at $P(x_0, y_0)$ is $\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$. For $P(2\cos\theta, \sqrt{3}\sin\theta)$, the tangent is $\frac{x(2\cos\theta)}{4} + \frac{y(\sqrt{3}\sin\theta)}{3} = 1$, which simplifies to $\frac{x\cos\theta}{2} + \frac{y\sin\theta}{\sqrt{3}} = 1$. The x-intercept (when $y=0$) is $A = \frac{2}{\cos\theta}$. The y-intercept (when $x=0$) is $B = \frac{\sqrt{3}}{\sin\theta}$. The length $L$ of the intercepted portion is $\sqrt{A^2+B^2} = \sqrt{\frac{4}{\cos^2\theta} + \frac{3}{\sin^2\theta}}$. To minimize $L$, we minimize $L^2 = \frac{4}{\cos^2\theta} + \frac{3}{\sin^2\theta}$. Let $u = \cos^2\theta$. $L^2 = \frac{4}{u} + \frac{3}{1-u}$. The minimum value of $L^2$ is $(\sqrt{4} + \sqrt{3})^2 = (2+\sqrt{3})^2$ when $\frac{4}{u} = \frac{3}{1-u}$, which yields $u=\frac{4}{7}$. The minimum length $L = \sqrt{(2+\sqrt{3})^2} = 2+\sqrt{3}$. This matches with option (2).

(S) If foot of normal from P on major axis is P' and normal at P cuts transverse axis at N, then $\frac{CN}{CP'}$ is Let $P(x_0, y_0) = (a\cos\theta, b\sin\theta)$. The foot of the perpendicular from P to the major axis is $P'=(x_0, 0)$. So, $P'=(2\cos\theta, 0)$. $CP' = |2\cos\theta|$. The equation of the normal at $P(x_0, y_0)$ is $\frac{ax}{\cos\theta} - \frac{by}{\sin\theta} = a^2-b^2$. The normal cuts the transverse axis (x-axis) at N. Setting $y=0$: $\frac{ax_N}{\cos\theta} = a^2-b^2 \implies x_N = \frac{(a^2-b^2)\cos\theta}{a}$. $a=2, b=\sqrt{3}, a^2=4, b^2=3$. $x_N = \frac{(4-3)\cos\theta}{2} = \frac{1}{2}\cos\theta$. So $N = (\frac{1}{2}\cos\theta, 0)$. $CN = |\frac{1}{2}\cos\theta|$. Then $\frac{CN}{CP'} = \frac{|\frac{1}{2}\cos\theta|}{|2\cos\theta|} = \frac{1/2}{2} = \frac{1}{4}$. This matches with option (4).

The correct matches are P-3, Q-1, R-2, S-4.