Question

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

• A. \(\dfrac{x^2}{4a^2}-\dfrac{y^2}{4b^2}=1\)
• B. \(\left(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}\right)=\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}\)
• C. \(\dfrac{x}{a}-\dfrac{y}{b}=\dfrac1{a^2}+\dfrac1{b^2}\)
• D. \(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}=\dfrac1{a^2}-\dfrac1{b^2}\)

Hint:

For a pair of homogeneous lines \(Ax^2+2Hxy+By^2=0\), the lines are perpendicular when \(A+B=0\).

Answer 1

The Correct Option is D

Step 1: Write the equation of chord of contact.
Let the external point be \((x_1,y_1)\).
For the hyperbola
\[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the chord of contact from \((x_1,y_1)\) is given by
\[ \frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1 \]

Step 2: Find the intercept form of the chord.
Rewrite the equation as
\[ \frac{x}{a^2/x_1}-\frac{y}{b^2/y_1}=1 \]
Thus, the line cuts the coordinate axes at
\[ \left(\frac{a^2}{x_1},0\right) \] and
\[ \left(0,-\frac{b^2}{y_1}\right) \]
These two points form the chord of contact.

Step 3: Use the condition that the chord subtends a right angle at the origin.
If a line segment subtends a right angle at the origin, then the vectors joining the origin to its endpoints are perpendicular.
Hence,
\[ \left(\frac{a^2}{x_1},0\right)\cdot \left(0,-\frac{b^2}{y_1}\right)=0 \] is not directly useful because axes intercepts are perpendicular already.
Instead, use the condition for the pair of tangents from the origin to be perpendicular.
For the chord of contact
\[ T=0, \] the combined equation of tangents from the origin is obtained by homogenizing:
\[ \left(\frac{xx_1}{a^2}-\frac{yy_1}{b^2}\right)^2 = \frac{x^2}{a^2}-\frac{y^2}{b^2} \]

Step 4: Use the condition for perpendicular pair of lines.
For a homogeneous second-degree equation
\[ Ax^2+2Hxy+By^2=0, \] the lines are perpendicular if
\[ A+B=0 \]
Expanding, we get
\[ \left(\frac{x_1^2}{a^4}-\frac1{a^2}\right)x^2 -\frac{2x_1y_1}{a^2b^2}xy + \left(\frac{y_1^2}{b^4}+\frac1{b^2}\right)y^2 =0 \]
Thus,
\[ A=\frac{x_1^2}{a^4}-\frac1{a^2} \] and
\[ B=\frac{y_1^2}{b^4}+\frac1{b^2} \]
Using \(A+B=0\),
\[ \frac{x_1^2}{a^4}-\frac1{a^2} + \frac{y_1^2}{b^4} +\frac1{b^2} =0 \]
Rearranging,
\[ \frac{x_1^2}{a^4} + \frac{y_1^2}{b^4} = \frac1{a^2}-\frac1{b^2} \]

Step 5: Write the locus.
Replacing \((x_1,y_1)\) by \((x,y)\), the locus is
\[ \boxed{ \frac{x^2}{a^4}+\frac{y^2}{b^4} = \frac1{a^2}-\frac1{b^2} } \]