Question

Prove x²=4y,x²+4y=8 intersect orthogonally at (2,1) and (-2,1).

Answer 1

we have to prove that, x² = 4y and x² + 4y = 8 intersect orthogonally or right angles at (2,1) and (-2,1).

proof : first differentiate the curve, x² = 4y with respect to x,

i.e., 2x = 4 dy/dx ⇒dy/dx = x/2

at (2, 1) , slope of tangent of curve, x² = 4y is $\frac{dy}{dx}|_{(2,1)}$ = (2)/2 = 1

and slope of tangent at (-2, 1) is $\frac{dy}{dx}|_{(-2,1)}$ is (-2)/2 = -1

similarly, difference the curve, x² + 4y = 8 with respect to x,

i.e., 2x + 4 dy/dx = 0 ⇒dy/dx = -x/2

at (2,1), slope of tangent of the curve, x² + 4y = 8 is $\frac{dy}{dx}|_{(2,1)}$ = -(2)/2 = -1

slope of tangent at (-2, 1) is $\frac{dy}{dx}|_{(-2,1)}$ = -(-2)/2 = 1

here we see,

slope of tangent of first curve at (2,1) × slope of tangent of 2nd curve at (2,1) = -1

hence, curves intersect orthogonally at (2,1)

again we see,

slope of tangent of first curve at (-2,1) × slope of tangent of 2nd curve at (-2,1) = -1

hence, curves also intersect orthogonally at (-2,1)