Question

Verify that the point P(acosθ,bsinθ) lies on the ellipse
x^2/ a^2+ y^2/ b^2= 1,
where a and b are the semi-major and semi-minor axes respectively of the ellipse . Find the gradient of the tangent to the curve at P and show that the equation of the normal at P is axsinθ−by cosθ = (a^2 −b^2)sinθ cosθ. If P is not on the axes and if the normal at P passes through the point B(0,b), Show that a^2 > 2b^2. If further, the tangent at P meets the y-axis at Q, show that
|BQ| =
a^2/ b^2
.

Answer 1

equation of ellipse x²/a² + y²/b² = 1

we have points P(acos$\theta$ + bsin$\theta$ )

put the points in the equation of the ellipse

take L.H.S

(acos$\theta$)²/a² + (bsin$\theta$)²/b²

a²cos²$\theta$ /a² + b²sin² $\theta$/b²

cos² $\theta$ + sin²$\theta$

=1

hence the points are lies in the ellipse

the equation of normal is at p is ax sin$\theta$ - by cos$\theta$ = (a²-b²)sin$\theta$ cos$\theta$

the gradient of the equtation is dy/dx

so the gradient is

dy/dx( ax sin$\theta$) - dy/dx(by cos$\theta$ )= dy/dx {(a²-b²)sin$\theta$cos$\theta$}

asin$\theta$ -0=0 is the gradient of the tangent to the curve at P

equation of the normal is

axsin$\theta$ -bycos$\theta$ = (a²-b²)sin$\theta$ cos$\theta$ at point B(0,b)

a(0)sin$\theta$ - b(b)cos$\theta$ = (a²-b²)sin$\theta$ cos$\theta$

0-b²cos$\theta$ =(a²-b²)sin$\theta$ cos$\theta$

hence a²>2b²

if the tangent P meets the y axis at Q

than |BQ| = a²/b²

the point B (0,b) similarly at y axis the point Q (a,0)

so the the point BQ(a,b)

hence |BQ| = a²/b²