The locus of a variable point whose chord of | vedclass

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

Question

(D) The equation of the given hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Let $(h, k)$ be the pole of a chord $PQ$ of the hyperbola. The equatio

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The equation of the given hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Let $(h, k)$ be the pole of a chord $PQ$ of the hyperbola. The equation of the chord of contact $PQ$ is $\frac{xh}{a^2}-\frac{yk}{b^2}=1$. To find the combined equation of the lines joining the origin to the points of intersection of the hyperbola and the chord,we homogenize the hyperbola equation using the chord equation: $\frac{x^2}{a^2}-\frac{y^2}{b^2} = \left(\frac{xh}{a^2}-\frac{yk}{b^2}\right)^2$. Since the chord $PQ$ subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero. Expanding the right side: $\frac{x^2}{a^2}-\frac{y^2}{b^2} = \frac{x^2h^2}{a^4} + \frac{y^2k^2}{b^4} - \frac{2xyhk}{a^2b^2}$. Rearranging terms: $x^2\left(\frac{1}{a^2}-\frac{h^2}{a^4}\right) + y^2\left(-\frac{1}{b^2}-\frac{k^2}{b^4}\right) + \frac{2xyhk}{a^2b^2} = 0$. Setting the sum of coefficients of $x^2$ and $y^2$ to zero: $\left(\frac{1}{a^2}-\frac{h^2}{a^4}\right) + \left(-\frac{1}{b^2}-\frac{k^2}{b^4}\right) = 0$. $\Rightarrow \frac{h^2}{a^4} + \frac{k^2}{b^4} = \frac{1}{a^2} - \frac{1}{b^2}$. Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{a^4} + \frac{y^2}{b^4} = \frac{1}{a^2} - \frac{1}{b^2}$.

The locus of a variable point whose chord of contact with respect to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends a right angle at the origin is

Similar Questions

Find the equation of the hyperbola whose foci are $(6, 5)$ and $(-4, 5)$ and eccentricity is $5/4$.

Find the condition for the curves $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $xy = c^2$ to intersect orthogonally.

Find the equations of the tangent and normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at the point $(x_{0}, y_{0})$.

The eccentricity of the hyperbola $2x^2 - y^2 = 6$ is

The equation of the hyperbola in the standard form (with transverse axis along the $x$-axis) having the length of the latus rectum = $9$ units and eccentricity = $5/4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module