Search any question & find its solution
Question:
Answered & Verified by Expert
If $p, q$ are the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola respectively, then the area of the square (in sq. units) formed by the points of intersection of the ellipse $\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$ and the pair of lines $x^2-y^2=0$ is
Options:
Solution:
1691 Upvotes
Verified Answer
The correct answer is:
4
$p$ and $q$ are eccentricity of hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate of hyperbola,
$\therefore \quad p^2=\frac{a^2+b^2}{a^2} \text { and } q^2=\frac{a^2+b^2}{b^2}$
Solving Eqs. (i) and (ii), we get
$(1,1),(-1,-1),(-1,1)(1,-1)$
Area of square $A B C D,=A B^2=(2)^2=4$
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate of hyperbola,
$\therefore \quad p^2=\frac{a^2+b^2}{a^2} \text { and } q^2=\frac{a^2+b^2}{b^2}$
Solving Eqs. (i) and (ii), we get
$(1,1),(-1,-1),(-1,1)(1,-1)$
Area of square $A B C D,=A B^2=(2)^2=4$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.