Question

If the equations of the tangents at P.Q.
and the vertex A of a parabola are
3x+4y-7=0, 2x+3y-10=0 and x-y=0​

Answer 1

Complete question:

If equation of tangent at P, Q and vertex A of a parabola are 3x + 4y ­ 7 = 0,  2x + 3y ­ 10 = 0 and x ­ y = 0 respectively, then

Option 1: Focus is (4, 5)

Option 2: Length of latus rectum is 2√2

Option 3: Axis is x + y - 9

Option 4: Vertex is (9/2, 9/2)

Answer:

All the options are applicable for the given question.

Step-by-step explanation:

The equation of parabola is given as:

((x + y + z)/√2)² = k((x - y)/√2)

From question, 3x + 4y - 7 = 0 is a tangent.

Now,

$\left(\frac{\frac{7-4 y}{3}+y+c}{\sqrt{2}}\right)^{2}=k\left(\frac{\frac{7-4 y}{3}-y}{\sqrt{2}}\right)$

The above formed equation is quadratic equation.

Δ = 0 ⇒ (21√2k - 14 - 6c)² = 4[(7 + 3c)² - 21√2k] → (equation 1)

Now, for the tangent 2x + 3y - 10 = 0, we get,

$\left(\frac{\frac{10-3y}{2}+y+c}{\sqrt{2}}\right)^{2}=k\left(\frac{\frac{10-3y}{2}-y}{\sqrt{2}}\right)$

Δ = 0 ⇒ (10√2k - 20 - 4c)² = 4[(10 + 2c)² - 20√2k] → (equation 2)

On solving equation (1) and (2), we get,

k = -2√2 and c = -9