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Question: A parabola touches straight lines $x - 2y + 13 = 0$ and $2x + y + 1 = 0$ at points $P(17, 15)$ and $...

A parabola touches straight lines x−2y+13=0x - 2y + 13 = 0 and 2x+y+1=02x + y + 1 = 0 at points P(17,15)P(17, 15) and Q(2,−5)Q(2, -5) respectively then:

A

Focus of parabola is (73,−3)(\frac{7}{3}, -3)

B

Directrix of parabola is x+y−2=0x + y - 2 = 0

C

Focus of parabola is (5,−1)(5, -1)

D

Length of latus rectum of parabola is 1616

Answer

C. Focus of parabola is (5,−1)(5, -1) and D. Length of latus rectum of parabola is 1616

Explanation

Solution

Here's how to solve this problem:

1. Find the intersection point R of the two tangents:

Solve the system of equations:

  • x−2y+13=0x - 2y + 13 = 0
  • 2x+y+1=02x + y + 1 = 0

This gives R(−3,5)R(-3, 5).

2. Use the property that the line joining the point of intersection of two tangents to the focus is perpendicular to the chord of contact PQ:

  • Slope of PQ=15−(−5)17−2=43PQ = \frac{15 - (-5)}{17 - 2} = \frac{4}{3}
  • Let S(α,β)S(\alpha, \beta) be the focus. Slope of RS=β−5α+3RS = \frac{\beta - 5}{\alpha + 3}
  • Since RS⊥PQRS \perp PQ, β−5α+3⋅43=−1  ⟹  3α+4β−11=0\frac{\beta - 5}{\alpha + 3} \cdot \frac{4}{3} = -1 \implies 3\alpha + 4\beta - 11 = 0

3. Use the property that the image of the focus with respect to any tangent lies on the directrix:

  • Image of SS w.r.t tangent L1L_1: x−2y+13=0x - 2y + 13 = 0 is S1S_1.
  • Image of SS w.r.t tangent L2L_2: 2x+y+1=02x + y + 1 = 0 is S2S_2.
  • Using the image formula, find S1S_1 and S2S_2. Substitute 3α+4β=113\alpha + 4\beta = 11 to simplify the x-coordinates. You'll find that both S1S_1 and S2S_2 have an x-coordinate of -3.

This means the directrix is the vertical line x=−3x = -3.

4. Find the coordinates of the focus S(α, β):

  • Use the definition of a parabola: distance from a point on the parabola to the focus equals its distance to the directrix.
  • For point P(17,15)P(17, 15): (α−17)2+(β−15)2=∣17−(−3)∣=20\sqrt{(\alpha - 17)^2 + (\beta - 15)^2} = |17 - (-3)| = 20
  • For point Q(2,−5)Q(2, -5): (α−2)2+(β+5)2=∣2−(−3)∣=5\sqrt{(\alpha - 2)^2 + (\beta + 5)^2} = |2 - (-3)| = 5

5. Solve the system of equations:

  • Solve the system formed by 3α+4β−11=03\alpha + 4\beta - 11 = 0 and (α−2)2+(β+5)2=25(\alpha - 2)^2 + (\beta + 5)^2 = 25. You will find that α=5\alpha = 5 and β=−1\beta = -1.
  • Therefore, the focus is (5,−1)(5, -1).

6. Calculate the length of the latus rectum:

  • The distance from the focus (5,−1)(5, -1) to the directrix x=−3x = -3 is 8.
  • The length of the latus rectum is 4a4a, and the distance from the focus to the directrix is 2a2a. Therefore, 2a=82a = 8, so a=4a = 4.
  • The length of the latus rectum is 4∗4=164 * 4 = 16.

Final Answer:

The correct options are:

  • C. Focus of parabola is (5,−1)(5, -1)
  • D. Length of latus rectum of parabola is 1616