Let the normals at all the points


Let the normals at all the points on a given curve pass through a fixed point $(a, b)$. If the curve passes through $(3,-3)$ and $(4,-2 \sqrt{2})$, and given that $a-2 \sqrt{2} b=3$, then $\left(a^{2}+b^{2}+a b\right)$ is equal to


All normals of circle passes through centre

Radius $=\mathrm{CA}=\mathrm{CB}$



$=(a-4)^{2}+(b-2 \sqrt{2})^{2}$

$a+(3-2 \sqrt{2}) b=3$

$a-2 \sqrt{2} b+3 b=3$...(1)

given that $a-2 \sqrt{2} b=3$...(2)

from (1) \& (2) $\Rightarrow a=3, b=0$

$a^{2}+b^{2}+a b=9$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now