Question

Prove x²=y and x³+6y=7 intersect at right angles at (1,1).

Answer 1

we have to prove that, x² = y and x³ + 6y = 7 intersect at right angles at (1,1).

proof : for curve, x² = y

differentiating with respect to x,

2x = dy/dx

or, dy/dx = 2x

at (1, 1) , slope of tangent of curve, x² = y is $m_1=\frac{dy}{dx}|_{(1,1)}$= 2 × 1 = 2

for curve, x³ + 6y = 7

differentiating both sides,

3x² + 6 dy/dx = 0

⇒dy/dx = -x²/2

at (1,1) slope of tangent of the curve, x³ + 6y = 7 is $m_2=\frac{dy}{dx}|_{(1,1)}$ = -(1)²/2 = -1/2

now, slope of tangent of first curve × slope of tangent of 2nd curve = 2 × -1/2 = -1

hence, tangent are perpendicular to each other at (1,1) and hence, curves intersect at right angle at (1,1)