The locus of a point whose chord of contact w | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The foci of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $(\pm ae, 0)$,where $e^2 = 1 + \frac{b^2}{a^2}$,so $a^2e^2 = a^2+b^2$.The circle

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The foci of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $(\pm ae, 0)$,where $e^2 = 1 + \frac{b^2}{a^2}$,so $a^2e^2 = a^2+b^2$.The circle described on the line joining the foci as diameter is $(x-ae)(x+ae) + y^2 = 0$,which simplifies to $x^2 + y^2 = a^2e^2 = a^2+b^2$.Let the point be $P(x_1, y_1)$. The equation of the chord of contact is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$,or $b^2x_1x - a^2y_1y - a^2b^2 = 0$.This line touches the circle $x^2 + y^2 = a^2+b^2$ if the perpendicular distance from the center $(0,0)$ to the line equals the radius $\sqrt{a^2+b^2}$.$\frac{|-a^2b^2|}{\sqrt{(b^2x_1)^2 + (-a^2y_1)^2}} = \sqrt{a^2+b^2}$.Squaring both sides: $\frac{a^4b^4}{b^4x_1^2 + a^4y_1^2} = a^2+b^2$.Rearranging gives $b^4x_1^2 + a^4y_1^2 = \frac{a^4b^4}{a^2+b^2}$.Dividing by $a^4b^4$,we get $\frac{x_1^2}{a^4} + \frac{y_1^2}{b^4} = \frac{1}{a^2+b^2}$.Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $\frac{x^2}{a^4} + \frac{y^2}{b^4} = \frac{1}{a^2+b^2}$.

The locus of a point whose chord of contact with respect to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ touches the circle described on the straight line joining the foci of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ as diameter is

Similar Questions

If the product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ of eccentricity $e = \sqrt{3}$ from its asymptotes is equal to $6$,then the length of the transverse axis of the hyperbola is

The product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt{2}$. Its equation is

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

Let $P(4,3)$ be a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. If the normal at $P$ intersects the $X$-axis at $(16,0)$,then the eccentricity of the hyperbola is

Vedclass Test Series

Exam Paper Generator

Online Exam Module