Answer to Question #124548 in Geometry for Max

Question

2. Verify that the point P(a cos θ, b sin θ) 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 thegradient of the tangent to the curve at P and show that the equation of the normal at P is
ax sin θ − by cos θ = (a^2 − b^2) sin θ cos θ.

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 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 these point are lies on 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 equation is dy/dx

so the gradient is

a sin$\theta$ {dy/dx(x)} - by cos$\theta$ {dy/dx(1)} =(a²-b²)sin$\theta$ cos$\theta$ {dy/dx(1)}

asin$\theta$ - 0 = 0

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

 

equation of normal is

ax sin$\theta$ - by cos$\theta$ = (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