# Equation of Normal to Ellipse in all Forms

## Equation of Normal to ellipse : $$x^2\over a^2$$ + $$y^2\over b^2$$ = 1

### (a) Point form :

The Equation of normal to the given ellipse at ($$x_1, y_1$$) is

$$a^2x\over x_1$$ + $$b^2y\over y_1$$ = $$a^2-b^2$$ = $$a^2e^2$$

Example : Find the normal to the ellipse $$9x^2+16y^2$$ = 288 at the point (4,3).

Solution : We have, $$9x^2+16y^2$$ = 288

Comparing with general equation of ellipse,

$$a^2$$ = 32 and $$b^2$$ = 18

The normal to given ellipse in point form is $$a^2x\over x_1$$ + $$b^2y\over y_1$$ = $$a^2-b^2$$

So, $$32x\over 4$$ + $$18y\over 3$$ = 14

8x + 6y = 14

Hence, normal to given ellipse is 8x + 6y = 14.

### (b) Slope form :

The equation of normal to the given ellipse whose slope is ‘m’, is

y = mx $$\mp$$ $${(a^2-b^2)m}\over \sqrt{a^2 + b^2m^2}$$

Example : Find the normal to the ellipse $$x^2 + 2y^2$$ = 6 whose slope is 2.

Solution : We have, $$x^2 + 2y^2$$ = 6

Comparing with $$x^2\over a^2$$ + $$y^2\over b^2$$ = 1

$$a^2$$ = 6 and $$b^2$$ = 3

The normal to given ellipse in slope form is y = mx $$\mp$$ $${(a^2-b^2)m}\over \sqrt{a^2 + b^2m^2}$$

So, y = 2x $$\mp$$ $${3m}\over \sqrt{18}$$

Hence, normal to given ellipse isy = 2x $$\mp$$ $${3m}\over \sqrt{18}$$.

### (c) Parametric form :

The equation of normal to the given ellipse at its point (acos$$\theta$$, bsin$$\theta$$), is

$$axsec\theta$$ – $$bycosec\theta$$ = $$a^2-b^2$$

or $$ax\over cos\theta$$ – $$by\over sin\theta$$ = $$a^2-b^2$$

Example : Find the condition that the line lx + my = n may be a normal to the ellipse $$x^2\over a^2$$ + $$y^2\over b^2$$ = 1

Solution : Equation of normal to the given ellipse at its point (acos$$\theta$$, bsin$$\theta$$), is

$$ax\over cos\theta$$ – $$by\over sin\theta$$ = $$a^2-b^2$$ ….(i)

If line lx + my = n is also normal to the ellipse then there must be a value of $$\theta$$ for which line (i) and line lx + my = n are identical. for that value of $$\theta$$ we have

$$l\over ({a\over cos\theta})$$ = $$m\over -({b\over sin\theta})$$ = $$n\over {l(a^2 – b^2)}$$ or

$$\cos\theta$$ = $$an\over {l(a^2 – b^2)}$$ ……(iii)

$$\sin\theta$$ = $$-bn\over {m(a^2 – b^2)}$$ ……(iv)

Squaring and adding (iii) and (iv), we get $$n^2\over {(a^2 – b^2)^2}$$ ($${a^2\over l^2} + {b^2\over m^2}$$) which is the required condition.

Hope you learnt equation of normal to ellipse in all forms, learn more concepts of ellipse and practice more questions to get ahead in the competition. Good luck!