Question

Consider the ellipse \(\frac{x^2}{9}+\frac{y^2}{4}=1\). Let S (p, q) be a point in the first quadrant such that \(\frac{p^2}{9}+\frac{q^2}{4}>1\). Two tangents are drawn from S to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point T in the fourth quadrant. Let R be the vertex of the ellipse with positive x-coordinate and Obe the center of the ellipse. If the area of the triangle △ORT is \(\frac{3}{2}\) , then which of the following options is correct ?

• A. \(q=2,p=3\sqrt3\)
• B. \(q=2,p=4\sqrt3\)
• C. \(0.5m,-5kg \ \frac{m}{s}\)
• D. \(-1m,5kg \ \frac{m}{s}\)

Answer 1

The Correct Option is A

To solve the problem, we analyze the ellipse and the tangents drawn from point \( S(p,q) \), and use the given conditions to find the required relationship.

1. Given ellipse equation:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \] with vertices:
- \( R = (3, 0) \) (vertex on positive x-axis)
- Minor axis endpoints: \( (0, 2) \) and \( (0, -2) \)
Center \( O = (0, 0) \).

2. Condition on point \( S(p,q) \):
\[ \frac{p^2}{9} + \frac{q^2}{4} > 1 \] So \( S \) lies outside the ellipse in the first quadrant, \( p > 0, q > 0 \).

3. Tangents from \( S \) touch the ellipse at two points:
- One tangent touches the ellipse at the endpoint of the minor axis: \( (0, 2) \) (since it is in the first quadrant)
- The other tangent touches the ellipse at \( T \) in the fourth quadrant (so \( y_T < 0 \)).

4. Equation of tangent to ellipse at a point \( (x_1, y_1) \) on ellipse:
\[ \frac{x x_1}{9} + \frac{y y_1}{4} = 1 \]

5. Tangent at \( (0, 2) \):
\[ \frac{x \cdot 0}{9} + \frac{y \cdot 2}{4} = 1 \implies \frac{2y}{4} = 1 \implies y = 2 \] This tangent is the horizontal line \( y = 2 \).

6. Since \( S(p,q) \) lies on the tangent, substitute \( y = 2 \) to find the point \( S \):
The tangent line \( y = 2 \) passes through \( S \), so \( q = 2 \).

7. Equation of tangent at point \( T(x_1, y_1) \) on ellipse:
\[ \frac{p x_1}{9} + \frac{q y_1}{4} = 1 \] But we don't know \( (x_1, y_1) = T \) yet, only that \( T \) is on the ellipse and in the fourth quadrant.

8. Use the tangent condition for \( S(p,q) \) to find the slope form:
Equation of tangent from external point \( S(p,q) \) to ellipse is:
\[ y = m x + c \] which passes through \( S \) and is tangent to the ellipse.
Or better, use the polar form of tangent from point \( (p,q) \):

9. Using the general tangent equation from \( S(p,q) \) to ellipse:
The length of tangent from \( S \) to ellipse is given by:
\[ \frac{p^2}{9} + \frac{q^2}{4} > 1 \] and the equation of tangent from point \( S \) to ellipse can be written as:
\[ y = m x + c \] with \[ c = q - m p \]

10. Condition for tangency:
Substitute \( y = m x + c \) into ellipse equation:
\[ \frac{x^2}{9} + \frac{(m x + c)^2}{4} = 1 \] This is a quadratic in \( x \). Tangent means discriminant \( = 0 \):

11. Write the quadratic in \( x \):
\[ \frac{x^2}{9} + \frac{m^2 x^2 + 2 m c x + c^2}{4} = 1 \] Multiply both sides by 36 (LCM of denominators 9 and 4):
\[ 4 x^2 + 9 m^2 x^2 + 18 m c x + 9 c^2 = 36 \] \[ (4 + 9 m^2) x^2 + 18 m c x + (9 c^2 - 36) = 0 \]

12. Discriminant \( D = 0 \) for tangent:
\[ D = (18 m c)^2 - 4 (4 + 9 m^2)(9 c^2 - 36) = 0 \] \[ 324 m^2 c^2 = 4 (4 + 9 m^2)(9 c^2 - 36) \]

13. Substitute \( c = q - m p = 2 - m p \):
Since \( q = 2 \), rewrite:
\[ c = 2 - m p \]

14. The second tangent touches the ellipse at \( T(x_T, y_T) \) in the fourth quadrant, so \( y_T < 0 \).

15. Calculate the area of triangle \( \triangle O R T \):
Points:
\[ O = (0,0), \quad R = (3, 0), \quad T = (x_T, y_T) \] Area formula: \[ \text{Area} = \frac{1}{2} \times |x_T \cdot (0 - 0) + 3 \cdot (y_T - 0) + 0 \cdot (0 - y_T)| = \frac{1}{2} \times |3 y_T| = \frac{3}{2} |y_T| \] Given area = \(\frac{3}{2}\), so: \[ \frac{3}{2} |y_T| = \frac{3}{2} \implies |y_T| = 1 \] Since \( y_T < 0 \) (fourth quadrant), \( y_T = -1 \).

16. Find \( x_T \) using ellipse equation:
\[ \frac{x_T^2}{9} + \frac{(-1)^2}{4} = 1 \implies \frac{x_T^2}{9} + \frac{1}{4} = 1 \implies \frac{x_T^2}{9} = \frac{3}{4} \implies x_T^2 = \frac{27}{4} \] \[ x_T = \frac{3 \sqrt{3}}{2} \quad \text{(positive x-coordinate in fourth quadrant)} \]

17. Equation of tangent at \( T \):
Using point \( T \) on ellipse:
\[ \frac{x x_T}{9} + \frac{y y_T}{4} = 1 \implies \frac{x \cdot \frac{3 \sqrt{3}}{2}}{9} + \frac{y \cdot (-1)}{4} = 1 \] Simplify: \[ \frac{x \sqrt{3}}{6} - \frac{y}{4} = 1 \] Rearranged: \[ y = \frac{2}{3} \sqrt{3} x - 4 \]

18. Since \( S(p, q) \) lies on the two tangents, \( S \) lies on both \( y=2 \) and the tangent through \( T \):
Substitute \( q = 2 \) in tangent through \( T \): \[ 2 = \frac{2}{3} \sqrt{3} p - 4 \implies \frac{2}{3} \sqrt{3} p = 6 \implies p = \frac{6 \times 3}{2 \sqrt{3}} = \frac{18}{2 \sqrt{3}} = \frac{9}{\sqrt{3}} = 3 \sqrt{3} \]

19. Check that \( S(p, q) = (3 \sqrt{3}, 2) \) lies outside ellipse:
\[ \frac{(3 \sqrt{3})^2}{9} + \frac{2^2}{4} = \frac{27}{9} + 1 = 3 + 1 = 4 > 1 \] Condition satisfied.

Final Answer:
\[ \boxed{ p = 3 \sqrt{3}, \quad q = 2 } \] and the point \( S \) is \( \left(3 \sqrt{3}, 2\right) \).

Answer 2

The equation of the ellipse is: 

\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \quad (1) \]

Given \( q = 2 \), the area of \( \triangle ORT \) is:

\[ \text{Area} = \frac{1}{2} \times (OR) \times (OK) = \frac{3}{2}. \] \[ \Rightarrow \frac{1}{2} \times 3 \times k = \frac{3}{2} \] \[ \Rightarrow k = 1. \]

Let \( T(h, -1) \) lie on the ellipse (1):

\[ \frac{h^2}{9} + \frac{1}{4} = 1 \] \[ \Rightarrow \frac{h^2}{9} = \frac{3}{4} \] \[ \Rightarrow h^2 = \frac{9}{4} \] \[ \Rightarrow h = \pm \frac{3\sqrt{3}}{2}. \]

The equation of the tangent at \( T \left( \frac{3\sqrt{3}}{2}, -1 \right) \) is:

\[ \frac{3\sqrt{3}}{2} \cdot \frac{x}{9} + \frac{-y}{4} = 1 \] \[ \Rightarrow \frac{\sqrt{3}}{6} x - \frac{y}{4} = 1. \]

This line passes through \( S(p, 2) \):

\[ \frac{\sqrt{3}}{6} p - \frac{2}{4} = 1 \] \[ \Rightarrow \frac{\sqrt{3}}{6} p = \frac{3}{2} \] \[ \Rightarrow p = \frac{9\sqrt{3}}{2} = 3\sqrt{3}. \]

Thus, \( p = 3\sqrt{3} \) and \( q = 2 \).

Final Answer:

\[ q = 2, \quad p = 3\sqrt{3} \]