Two tangents to an exponential-logarithmic curve

A curve CC is given by the equation ex2y=ln⁡x2e^{x^2}y=\ln x^2, where x≠0x\ne0.

(i) Show that 2xln⁡x2+ex2dydx=2x2x\ln x^2+e^{x^2}\frac{dy}{dx}=\frac2x. marks

(ii) Find the equation of the tangent to CC at the point PP where x=1x=1. Given that the tangent at PP meets CC again at the point QQ, find the coordinates of QQ, leaving your answer correct to 5 significant figures. marks

(iii) Determine the acute angle, in degrees, between the tangents to CC at the points PP and QQ. marks

[2022/HCI/Prelim/I/9]

Total marks: 9

Solution

If this part uses an earlier result, revisit (ii) tangent Tangent at a point or open all working together.

(ii) second intersection

Intersections by calculator

The tangent at PP meets CC again at QQ. Find the coordinates of QQ, correct to 5 significant figures.

Explore this idea

Hint

  1. Set ln⁡x2ex2=2e(x−1)\frac{\ln x^2}{e^{x^2}}=\frac2e(x-1).

  2. Solve with the graphing calculator and exclude x=1x=1.

Answer · (ii) second intersection

Q=(−0.49516,−1.1001)Q=(-0.49516,-1.1001) (5 s.f.).

Explore the idea

Open interactive exploration

The tangent at P cuts the curve again at Q

Jump to P and Q. The fixed purple line is the tangent at P.

Position along the curve(1, 0)
Try an example
Tangent gradient
0.73576

At P: gradient 2e≈0.736\frac2e\approx0.736. The tangent here is the purple line.

For finite nonzero gradients, tangent gradient × normal gradient = −1. Both axes use equal scales.

  • E^(x²) y = ln x²
  • Tangent at the moving point
Key idea

The tangent at PP is also a chord of the curve, meeting it again at QQ.

Ideas to try
Slide to QQ and compare the two tangent lines.