Question

The tangent to the parabola y² = 4x at the point where it intersects the circle x² + y² = 5 in the first quadrant, passes through the point:
1. \(\left( { - \frac{1}{3},{\rm{\;}}\frac{4}{3}} \right)\)
2. \(\left( {\frac{1}{4},\frac{3}{4}} \right)\)
3. \(\left( {\frac{3}{4},\frac{7}{4}} \right)\)
4. \(\left( { - \frac{1}{4},\frac{1}{2}} \right)\)

Answer 1

Correct Answer - Option 3 : \(\left( {\frac{3}{4},\frac{7}{4}} \right)\)

A secant of a parabola is a line, or line segment, that joins two distinct points on the parabola. A tangent is a line that touches the parabola at exactly one point.

To find intersection point of x² + y² = 5 and y² = 4x substitute y² = 4x in x² + y² = 5 we get

⇒ x² + 4x – 5 = 0

⇒ x² + 5x – x – 5 = 0

⇒ x(x + 5) – 1(x + 5) = 0

⇒ (x – 1)(x + 5) = 0

∴ x = 1, -5

∴ y² = 4x = 4(1) = 4

⇒ y = 2

Intersection point in 1^st quadrant be (1, 2).

Now, equation of tangent to y2 = 4x at (1, 2) is

y × 2 = 2(x + 1)

⇒ y = x + 1

⇒ x – y + 1 = 0     ----(1)

By substituting the values given in options in equation (1), we get

Option A, \(\left( { - \frac{1}{3},\;\frac{4}{3}} \right) \Rightarrow - \frac{1}{3} - \frac{4}{3} + 1 = - \frac{5}{3} + 1 \ne 0\) 

Option B, \(\left( {\frac{1}{4},\frac{3}{4}} \right) \Rightarrow \frac{1}{4} - \frac{3}{4} + 1 = - \frac{2}{4} + 1 \ne 0\) 

Option C, \(\left( {\frac{3}{4},\frac{7}{4}} \right) \Rightarrow \frac{3}{4} - \frac{7}{4} + 1 = - \frac{4}{4} + 1 = 0\) 

Option D, \(\left( { - \frac{1}{4},\frac{1}{2}} \right) \Rightarrow - \frac{1}{4} - \frac{1}{2} + 1 = - \frac{3}{4} + 1 \ne 0\) 

Hence, \(\left( {\frac{3}{4},\frac{7}{4}} \right)\) lies on (1)

Thus, option (c) is the correct answer.