From any point on the conic x2a2+y2b2=4, tangents are drawn to the conic x2a2+y2b2=1, then locus of the normal at the points of contact meet on the conic and the equation of the conic is

Equations of concentric ellipses are x2a2+y2b2=4……………(1) x2a2+y2b2=1……………(2)Points P(acosα,bsinα) and Q(acosβ,bsinβ) on the (2) ellipse. Equations of tangents at P and Q are xacosα+ybsinα=1………(3) xacosβ+ybsinβ=1 ………..(4)Point of intersection of (3) & (4) is A{acos(α+β2)cos(α−β2),bsin(α+β2)cos(α−β2)}This point A satisfy (1)∴cos2(α+β2)cos2(α−β2)+sin2(α+β2)cos2(α−β2)=4 ⇒1=4cos2(α−β2),12=1+cos(α−β) ∴cos(α−β)=−12 …………(5)Equations of normal PR & QR are axsecα−bycosecα=a2−b2 & axsecβ−bycosecβ=a2−b2Or axsinα−bycosα=(a2−b2)sinαcosα ……………(6) axsinβ−bycosβ=(a2−b2)sinβcosβ ……………(7)Multiplying (6) by cosβ & (7) by cosα and subtracting, we get axsin(α−β)=(a2−b2)cosαcosβ(sinα−sinβ)∴axa2−b2=cosαcosβ.2cos(α+β2)sin(α−β2)2sin(α−β2)cos(α−β2)=cosαcosβ.cos(α+β2)cos(α−β2) =±2cosαcosβ.cos(α+β2) (from 5) =±cos(α+β2){cos(α+β)+cos(α−β)} axa2−b2=±cos(α+β2){cos(α+β)−12} Using (5) ⇒a2x2(a2−b2)2=cos2(α+β2){cos(α+β)−12}2Similarly⇒b2y2(a2−b2)2=sin2(α+β2){cos(α+β)+12}2 a2x2(a2−b2)2+b2y2(a2−b2)2=cos2(α+β)+14−cos2(α+β)Hence the required locus is a2x2+b2y2=14(a2−b2)2

From any point on the conic x2a2+y2b2=4, tangents are drawn to the conic x2a2+y2b2=1, then locus of the normal at the points of contact meet on the conic and the equation of the conic is

Equations of concentric ellipses are x2a2+y2b2=4……………(1) x2a2+y2b2=1……………(2)Points P(acosα,bsinα) and Q(acosβ,bsinβ) on the (2) ellipse. Equations of tangents at P and Q are xacosα+ybsinα=1………(3) xacosβ+ybsinβ=1 ………..(4)Point of intersection of (3) & (4) is A{acos(α+β2)cos(α−β2),bsin(α+β2)cos(α−β2)}This point A satisfy (1)∴cos2(α+β2)cos2(α−β2)+sin2(α+β2)cos2(α−β2)=4 ⇒1=4cos2(α−β2),12=1+cos(α−β) ∴cos(α−β)=−12 …………(5)Equations of normal PR & QR are axsecα−bycosecα=a2−b2 & axsecβ−bycosecβ=a2−b2Or axsinα−bycosα=(a2−b2)sinαcosα ……………(6) axsinβ−bycosβ=(a2−b2)sinβcosβ ……………(7)Multiplying (6) by cosβ & (7) by cosα and subtracting, we get axsin(α−β)=(a2−b2)cosαcosβ(sinα−sinβ)∴axa2−b2=cosαcosβ.2cos(α+β2)sin(α−β2)2sin(α−β2)cos(α−β2)=cosαcosβ.cos(α+β2)cos(α−β2) =±2cosαcosβ.cos(α+β2) (from 5) =±cos(α+β2){cos(α+β)+cos(α−β)} axa2−b2=±cos(α+β2){cos(α+β)−12} Using (5) ⇒a2x2(a2−b2)2=cos2(α+β2){cos(α+β)−12}2Similarly⇒b2y2(a2−b2)2=sin2(α+β2){cos(α+β)+12}2 a2x2(a2−b2)2+b2y2(a2−b2)2=cos2(α+β)+14−cos2(α+β)Hence the required locus is a2x2+b2y2=14(a2−b2)2

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From any point on the conic $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 4$ , tangents are drawn to the conic $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , then locus of the normal at the points of contact meet on the conic and the equation of the conic is

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$a^{2} x^{2} + b^{2} y^{2} = \frac{1}{4} {(a^{2} + b^{2})}^{2}$

$a^{2} x^{2} - b^{2} y^{2} = 4 {(a^{2} - b^{2})}^{2}$

$a^{2} x^{2} + b^{2} y^{2} = 4 {(a^{2} - b^{2})}^{2}$

$a^{2} x^{2} + b^{2} y^{2} = \frac{1}{4} {(a^{2} - b^{2})}^{2}$

answer is B.

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Detailed Solution

Equations of concentric ellipses are
   $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 4$ ……………(1)
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ……………(2)
Points $P (a \cos α, b \sin α)$ and $Q (a \cos β, b \sin β)$ on the (2) ellipse. Equations of tangents at P and Q are
   $\frac{x}{a} \cos α + \frac{y}{b} \sin α = 1$ ………(3)
   $\frac{x}{a} \cos β + \frac{y}{b} \sin β = 1$ ………..(4)
Point of intersection of (3) & (4) is
Question Image

$A {\frac{a \cos (\frac{α + β}{2})}{\cos (\frac{α - β}{2})}, \frac{b \sin (\frac{α + β}{2})}{\cos (\frac{α - β}{2})}}$

This point A satisfy (1)

$∴ \frac{\cos^{2} (\frac{α + β}{2})}{\cos^{2} (\frac{α - β}{2})} + \frac{\sin^{2} (\frac{α + β}{2})}{\cos^{2} (\frac{α - β}{2})} = 4 \Rightarrow 1 = 4 \cos^{2} (\frac{α - β}{2}), \frac{1}{2} = 1 + \cos (α - β) ∴ \cos (α - β) = - \frac{1}{2} \dots \dots \dots \dots (5)$

Equations of normal PR & QR are $a x \sec α - b y \cos e c α = a^{2} - b^{2}$ & $a x \sec β - b y \cos e c β = a^{2} - b^{2}$

Or $a x \sin α - b y \cos α = (a^{2} - b^{2}) \sin α \cos α \dots \dots \dots \dots \dots (6) a x \sin β - b y \cos β = (a^{2} - b^{2}) \sin β \cos β \dots \dots \dots \dots \dots (7)$

Multiplying (6) by $\cos β$ & (7) by $\cos α$ and subtracting, we get

$a x \sin (α - β) = (a^{2} - b^{2}) \cos α \cos β (\sin α - \sin β)$

$∴ \frac{a x}{a^{2} - b^{2}} = \frac{\cos α \cos β .2 \cos (\frac{α + β}{2}) \sin (\frac{α - β}{2})}{2 \sin (\frac{α - β}{2}) \cos (\frac{α - β}{2})}$

$= \frac{\cos α \cos β . \cos (\frac{α + β}{2})}{\cos (\frac{α - β}{2})} = \pm 2 \cos α \cos β . \cos (\frac{α + β}{2}) (f r o m 5) = \pm \cos (\frac{α + β}{2}) {\cos (α + β) + \cos (α - β)} \frac{a x}{a^{2} - b^{2}} = \pm \cos (\frac{α + β}{2}) {\cos (α + β) - \frac{1}{2}} U \sin g (5) \Rightarrow \frac{a^{2} x^{2}}{{(a^{2} - b^{2})}^{2}} = \cos^{2} (\frac{α + β}{2}) {\cos (α + β) - \frac{1}{2}}^{2}$

Similarly

$\Rightarrow \frac{b^{2} y^{2}}{{(a^{2} - b^{2})}^{2}} = \sin^{2} (\frac{α + β}{2}) {\cos (α + β) + \frac{1}{2}}^{2} \frac{a^{2} x^{2}}{{(a^{2} - b^{2})}^{2}} + \frac{b^{2} y^{2}}{{(a^{2} - b^{2})}^{2}} = \cos^{2} (α + β) + \frac{1}{4} - \cos^{2} (α + β)$