Start learning →

Which of the following is the slope form of a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$?

A$y=mx\pm\sqrt{a^2m^2+b^2}$
B$y=mx\pm\sqrt{a^2+b^2m^2}$
C$y=mx\pm\sqrt{a^2m^2-b^2}$
D$y=mx\pm\sqrt{b^2m^2-a^2}$
Answer & Solution
Correct answer: A. $y=mx\pm\sqrt{a^2m^2+b^2}$
Take the line $y=mx+c$. For tangency with the ellipse, substituting into $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ must give a quadratic in $x$ with discriminant zero. This leads to $c^2=a^2m^2+b^2$, so $c=\pm\sqrt{a^2m^2+b^2}$. **Hence the tangent in slope form is $y=mx\pm\sqrt{a^2m^2+b^2}$.**
Practise this question with Quill — JEE Advanced →

Keep practising Straight Lines and Conic Sections

Browse all Straight Lines and Conic Sections questions for JEE Advanced, or explore more Mathematics topics.

Start a Straight Lines and Conic Sections lesson →