Question

The point of contact of the tangent to the parabola $y^2 = 9x$ which passes through the point $(4, 10)$ and makes an angle $\theta$ with the positive side of the axis of the parabola, where $\tan \theta>2$, is

• A. (4/9, 2)
• B. (4, 6)
• C. (4, 5)
• D. (1/4, 1/6)

Answer 1

The Correct Option is A

The given parabola is: \[ y^2 = 9x \] This is of the standard form \( y^2 = 4ax \) where \( 4a = 9 \Rightarrow a = \frac{9}{4} \)

The equation of the tangent to the parabola \( y^2 = 4ax \) at point \( (x_1, y_1) \) is: \[ y y_1 = 2a(x + x_1) \]

We are told that the tangent passes through the external point \( (4, 10) \), and makes an angle \( \theta \) with the x-axis such that \( \tan \theta > 2 \).

So, we are to find the point on the parabola \( (x_1, y_1) \) such that the tangent from there:

1. Satisfies \( y_1^2 = 9x_1 \) (since the point lies on the parabola)
2. Passes through \( (4, 10) \)
3. Has slope \( m \) such that \( \tan \theta = m > 2 \)

Alternate method: Let us assume general equation of tangent in slope form for the parabola \( y^2 = 4ax \): \[ y = mx + \frac{a}{m} \Rightarrow \text{Here, } y = mx + \frac{9}{4m} \]

This tangent must pass through \( (4, 10) \). So substitute: \[ 10 = 4m + \frac{9}{4m} \] Multiply both sides by \( 4m \): \[ 40m = 16m^2 + 9 \Rightarrow 16m^2 - 40m + 9 = 0 \]

Solve the quadratic: \[ m = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 16 \cdot 9}}{2 \cdot 16} = \frac{40 \pm \sqrt{1600 - 576}}{32} = \frac{40 \pm \sqrt{1024}}{32} = \frac{40 \pm 32}{32} \]

So: \[ m = \frac{72}{32} = \frac{9}{4} \quad \text{or} \quad m = \frac{8}{32} = \frac{1}{4} \]

Now use the condition \( \tan \theta > 2 \). So choose: \[ m = \frac{9}{4} \]

Substitute this back into the slope form of tangent: \[ y = \frac{9}{4}x + \frac{9}{4 \cdot \frac{9}{4}} = \frac{9}{4}x + 1 \]

Find point of contact: For the parabola \( y^2 = 9x \), and the tangent \( y = \frac{9}{4}x + 1 \), find point of contact using standard method. Let the point be \( (x_1, y_1) \), then the tangent at \( (x_1, y_1) \) is: \[ y y_1 = 2a(x + x_1) = \frac{9}{2}(x + x_1) \Rightarrow y = \frac{9}{2y_1}(x + x_1) \]

But we already know that the tangent is: \[ y = \frac{9}{4}x + 1 \Rightarrow \text{So equating the two forms:} \] \[ \frac{9}{2y_1}(x + x_1) = \frac{9}{4}x + 1 \] Multiply both sides by \( 2y_1 \): \[ 9(x + x_1) = \frac{9}{2}x y_1 + 2y_1 \] Try a better way — we already have the slope \( m = \frac{9}{4} \), so plug into: \[ y = mx + \frac{a}{m} = \frac{9}{4}x + \frac{9}{4 \cdot \frac{9}{4}} = \frac{9}{4}x + 1 \] Let’s find point of contact on the parabola which lies on this tangent. Let point be \( (x_0, y_0) \) Then: \[ y_0^2 = 9x_0 \quad \text{and} \quad y_0 = \frac{9}{4}x_0 + 1 \] Substitute in first: \[ \left( \frac{9}{4}x_0 + 1 \right)^2 = 9x_0 \Rightarrow \frac{81}{16}x_0^2 + \frac{9}{2}x_0 + 1 = 9x_0 \Rightarrow \frac{81}{16}x_0^2 + \frac{9}{2}x_0 + 1 - 9x_0 = 0 \Rightarrow \frac{81}{16}x_0^2 - \frac{9}{2}x_0 + 1 = 0 \]

Multiply entire equation by 16: \[ 81x_0^2 - 72x_0 + 16 = 0 \] Solve: \[ x_0 = \frac{72 \pm \sqrt{(-72)^2 - 4 \cdot 81 \cdot 16}}{2 \cdot 81} = \frac{72 \pm \sqrt{5184 - 5184}}{162} = \frac{72}{162} = \frac{4}{9} \]

Now find \( y_0 \): \[ y_0 = \frac{9}{4} \cdot \frac{4}{9} + 1 = 1 + 1 = 2 \Rightarrow \text{Point of contact is } \left( \frac{4}{9}, 2 \right) \]

Answer:

\[ \boxed{\left( \frac{4}{9}, 2 \right)} \]