Question

The locus of mid point of normal chords of the ellipse $\frac{x^2}{a^2}+y^2{b}^2=1$

• A. $x^2+y^2=2$
• B. $x^2-y^2=1$
• C. $(\frac{a^6}{x^2}+\frac{b^6}{y^2}){(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2={(a^2-b^2)}^2$
• D. $x^2+y^2=1$

Answer 1

Answer: (C)

$(\frac{a^6}{x^2}+\frac{b^6}{y^2}){(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2={(a^2-b^2)}^2$

Let $P(acos\alpha ,bsin\alpha )$ and $Q(acos\beta ,bsin\beta )$ are two end-points of chord of ellipse.

Thus, equation of chord of ellipse passing through points P and Q having eccentric angles $\alpha$ and $\beta$ is given by,

$\frac{x}{a}cos\left(\frac{\alpha +\beta }{2}\right)+\frac{y}{b}sin\left(\frac{\alpha +\beta }{2}\right)=cos\left(\frac{\alpha -\beta }{2}\right)$ (1)

Let $(x_1,y_1)$ be coordinates of mid-point of chord PQ

$\therefore x_1=\frac{acos\alpha +acos\beta }{2}=\frac{a(cos\alpha +cos\beta )}{2}$

$\therefore x_1=\frac{a}{2}\left(2cos\left(\frac{\alpha +\beta }{2}\right)cos\left(\frac{\alpha -\beta }{2}\right)\right)$

$\therefore x_1=acos\left(\frac{\alpha +\beta }{2}\right)cos\left(\frac{\alpha -\beta }{2}\right)$

$y_1=\frac{bsin\alpha +bsin\beta }{2}=\frac{b(sin\alpha +sin\beta )}{2}$

$y_1=\frac{b}{2}\left(2sin\left(\frac{\alpha +\beta }{2}\right)cos\left(\frac{\alpha -\beta }{2}\right)\right)$

$y_1=bsin\left(\frac{\alpha +\beta }{2}\right)cos\left(\frac{\alpha -\beta }{2}\right)$ (2)

Equation of normal to the ellipse at $P(acos\alpha ,bsin\alpha )$ is given by,

$\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2-b^2$

$\therefore \frac{a^{2}x}{acos\alpha }+\frac{b^{2}y}{bsin\alpha }=a^2-b^2$

$\therefore \frac{x}{a}\left(\frac{a^2}{cos\alpha }\right)+\frac{y}{b}\left(\frac{b^2}{sin\alpha }\right)=a^2-b^2$ (3)

As chord is normal to the curve, equation (1) and (3) must be same. Thus equate the ratio of corresponding coefficients.

$\frac{cos\left(\frac{\alpha +\beta }{2}\right)}{\left(\frac{a^2}{cos\alpha }\right)}=\frac{sin\left(\frac{\alpha +\beta }{2}\right)}{\left(\frac{b^2}{sin\alpha }\right)}=\frac{cos\left(\frac{\alpha -\beta }{2}\right)}{a^2-b^2}$

$\frac{acos\left(\frac{\alpha +\beta }{2}\right)cos\left(\frac{\alpha -\beta }{2}\right)}{acos\left(\frac{\alpha -\beta }{2}\right)\left(\frac{a^2}{cos\alpha }\right)}=\frac{bsin\left(\frac{\alpha +\beta }{2}\right)cos\left(\frac{\alpha -\beta }{2}\right)}{bcos\left(\frac{\alpha -\beta }{2}\right)\left(\frac{b^2}{sin\alpha }\right)}=\frac{cos\left(\frac{\alpha -\beta }{2}\right)}{a^2-b^2}$

From equation (2),

$\frac{x_1}{cos\left(\frac{\alpha -\beta }{2}\right)\left(\frac{a^3}{cos\alpha }\right)}=\frac{y_1}{cos\left(\frac{\alpha -\beta }{2}\right)\left(\frac{b^3}{sin\alpha }\right)}=\frac{cos\left(\frac{\alpha -\beta }{2}\right)}{a^2-b^2}$

1) $\frac{x_1}{cos\left(\frac{\alpha -\beta }{2}\right)\left(\frac{a^3}{cos\alpha }\right)}=\frac{cos\left(\frac{\alpha -\beta }{2}\right)}{a^2-b^2}$

$\frac{x_{1}cos\alpha }{a^{3}cos\left(\frac{\alpha -\beta }{2}\right)}=\frac{cos\left(\frac{\alpha -\beta }{2}\right)}{a^2-b^2}$

$\therefore cos\alpha =\frac{a^3{cos}^2\left(\frac{\alpha -\beta }{2}\right)}{x_1(a^2-b^2)}$

2) $\frac{y_1}{cos\left(\frac{\alpha -\beta }{2}\right)\left(\frac{b^3}{sin\alpha }\right)}=\frac{cos\left(\frac{\alpha -\beta }{2}\right)}{a^2-b^2}$

$\therefore \frac{y_{1}sin\alpha }{b^{3}cos\left(\frac{\alpha -\beta }{2}\right)}=\frac{cos\left(\frac{\alpha -\beta }{2}\right)}{a^2-b^2}$

$\therefore sin\alpha =\frac{b^3{cos}^2\left(\frac{\alpha -\beta }{2}\right)}{y_1(a^2-b^2)}$

We know that ${sin}^2\alpha +{cos}^2\alpha =1$

$\therefore {\left(\frac{b^3{cos}^2\left(\frac{\alpha -\beta }{2}\right)}{y_1(a^2-b^2)}\right)}^2+{\left(\frac{a^3{cos}^2\left(\frac{\alpha -\beta }{2}\right)}{x_1(a^2-b^2)}\right)}^2=1$

$\therefore \frac{b^6{cos}^4\left(\frac{\alpha -\beta }{2}\right)}{y^2{(a^2-b^2)}^2}+\frac{a^6{cos}^4\left(\frac{\alpha -\beta }{2}\right)}{a^2{(a^2-b^2)}^2}=1$

$\therefore \frac{{cos}^4\left(\frac{\alpha -\beta }{2}\right)}{{(a^2-b^2)}^2}[\frac{b^6}{y^2}+\frac{a^6}{a^2}]=1$ (4)

From equation (2),

${\left(\frac{x_1}{a}\right)}^2+{\left(\frac{y_1}{b}\right)}^2={cos}^2\left(\frac{\alpha +\beta }{2}\right){cos}^2\left(\frac{\alpha -\beta }{2}\right)+{sin}^2\left(\frac{\alpha +\beta }{2}\right){cos}^2\left(\frac{\alpha -\beta }{2}\right)$

$\therefore {\left(\frac{x_1}{a}\right)}^2+{\left(\frac{y_1}{b}\right)}^2={cos}^2\left(\frac{\alpha -\beta }{2}\right)[{cos}^2\left(\frac{\alpha +\beta }{2}\right)+{sin}^2\left(\frac{\alpha +\beta }{2}\right)]$

$\therefore {\left(\frac{x_1}{a}\right)}^2+{\left(\frac{y_1}{b}\right)}^2={cos}^2\left(\frac{\alpha -\beta }{2}\right)$

Thus, from equation (4),

$\frac{{({\left(\frac{x_1}{a}\right)}^2+{\left(\frac{y_1}{b}\right)}^2)}^2}{{(a^2-b^2)}^2}[\frac{b^6}{y^2}+\frac{a^6}{a^2}]=1$

$\therefore [\frac{b^6}{y^2}+\frac{a^6}{a^2}]{({\left(\frac{x_1}{a}\right)}^2+{\left(\frac{y_1}{b}\right)}^2)}^2={(a^2-b^2)}^2$

$\therefore [\frac{b^6}{y^2}+\frac{a^6}{a^2}]{(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2={(a^2-b^2)}^2$