What is the solution?

Question

Let the tangents at point P and R on the parabola y = x² intersects at T. Tangent at point Q (lies in between the point P and R) on the parabola intersect PT and RT at A and B respectively.

what is the solution?

Answer 1

Given parabola is y = x²

The slope of tangent to given parabola is \(\frac{dy}{d\mathrm x}\) = 2x.

Let point

P(x₁, x₁²), Q(x₂, x₂²) & R(x₃, x₃²)

Equation of tangent PT is

y – x₁² = 2x₁(x – x₁)

⇒ y – 2x₁x + x₁² = 0 .........(1)

Equation of tangent RT is

y – x₃² = 2x₃(x – x₃)

⇒ y – 2x₃x + x₃² = 0 ..........(2)

Equation of tangent AB at Q is

y – x₂² = 2x₂(x – x₂)

⇒ y – 2x₂x + x₃² = 0 .........(3)

Now, subtract equation (2) from equation (1), we get

2x(x₃ – x₁) = x₃² – x₁²

⇒ 2x(x₃ – x₁) = (x₃ – x₁)(x₃ + x₁)   (\(\because\) a² – b² = (a + b)(a – b))

⇒ 2x = x₃ + x₁   (\(\because\) x₃ \(\ne\) x₁ because  P & T are different point)

⇒ x \(=\frac{\mathrm x_1+\mathrm x_3}{2}\)

Now, put \(\mathrm x = \frac{\mathrm x_1+\mathrm x_3}{2}\) in equation (1), we get

y = 2x₁\(\left(\frac{\mathrm x_1+\mathrm x_3}{2}\right)\) – x₁²

= x₁x₃

Thus, intersection point of equation (1) and (2) is \(\left(\frac{\mathrm x_1+\mathrm x_3}{2}, \mathrm x_1\mathrm x_3\right)\)

\(\because \) T is intersection point of tangents PT and RT.

\(\therefore\) T = \(\left(\frac{\mathrm x_1+\mathrm x_3}{2}, \mathrm x_1\mathrm x_3\right)\)

Similarly, A is intersection point of equation (1) and (3).

\(\therefore\) A = \(\left(\frac{\mathrm x_1+\mathrm x_2}{2}, \mathrm x_1, \mathrm x_2\right)\)

And B is intersection point of equation (2) and (3).

\(\therefore\) B = \(\left(\frac{\mathrm x_2+\mathrm x_3}{2}, \mathrm x_2\mathrm x_3\right)\)

Now, TA \(=\sqrt{\frac{(\mathrm x_2-\mathrm x_3)^2}{4}+\mathrm x_1^2(\mathrm x_2-\mathrm x_3)^2}\) \(=(\mathrm x_2-\mathrm x_3)\sqrt{\frac{1}{4}+\mathrm x_1^2}\)

TB \(=\sqrt{\frac{(\mathrm x_2-\mathrm x_1)^2}{4}+\mathrm x_3^2(\mathrm x_2-\mathrm x_1)^2}\) \(=(\mathrm x_2-\mathrm x_1)\sqrt{\frac{1}{4}+\mathrm x_3^2}\)

TP \(=\sqrt{\left(\mathrm x_1-\frac{\mathrm x_1+\mathrm x_3}{2}\right)^2+(\mathrm x_1^2-\mathrm x_1\mathrm x_3)^2}\)  (\(\because\) P = (x₁, x₁²))

\(=\sqrt{\left(\frac{\mathrm x_1-\mathrm x_3}{2}\right)^2+\mathrm x_1^2(\mathrm x_1-\mathrm x_3)^2}\)

\(=(\mathrm x_1-\mathrm x_3)\sqrt{\frac{1}{4}+\mathrm x_1^2}\)

TR \(=\sqrt{\left(\mathrm x_3-\frac{\mathrm x_1+\mathrm x_3}{2}\right)^2+(\mathrm x_3^2-\mathrm x_1\mathrm x_3)^2}\)  \(\mathrm{\left(\because R = (x_3, x_3^2\right)}\)

\(=\sqrt{\left(\frac{\mathrm x_3-\mathrm x_1}{2}\right)^2+\mathrm x_3^2(\mathrm x_3-\mathrm x_1)^2}\)

\(=(\mathrm x_3-\mathrm x_1)\sqrt{\frac{1}{4}+\mathrm x_3^2}\)

Now, \(\frac{TA}{TP}+\frac{TB}{TR}\) \(=\frac{(\mathrm x_2-\mathrm x_3)\sqrt{\frac{1}{4}+\mathrm x_1^2}}{(\mathrm x_1-\mathrm x_3)\sqrt{\frac{1}{4}+\mathrm x_1^2}}\) + \(\frac{(\mathrm x_2-\mathrm x_1)\sqrt{\frac{1}{4}+\mathrm x^2_3}}{(\mathrm x_3-\mathrm x_1)\sqrt{\frac{1}{4}+\mathrm x_3^2}}\)

\(=\frac{\mathrm x_2-\mathrm x_3}{\mathrm x_1-\mathrm x_3}+\frac{\mathrm x_2-\mathrm x_1}{\mathrm x_3-\mathrm x_1}\)

\(=\frac{\mathrm x_2-\mathrm x_3}{\mathrm x_1-\mathrm x_3}-\frac{\mathrm x_2-\mathrm x_1}{\mathrm x_1-\mathrm x_3}\)

\(=\frac{\mathrm x_2-\mathrm x_3-\mathrm x_2+\mathrm x_1}{\mathrm x_1-\mathrm x_3}\)

\(=\frac{\mathrm x_1-\mathrm x_3}{\mathrm x_1-\mathrm x_3}\)

= 1