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JEE Main 2021
Conic Sections
Parabola
Easy

Question

If the line ax + y = c, touches both the curves x 2 + y 2 = 1 and y 2 = 42\sqrt 2 x , then |c| is equal to :

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Solution

Key Concepts and Formulas for Common Tangents

To solve this problem, which asks for a line simultaneously tangent to a parabola and a circle, we employ two fundamental concepts from coordinate geometry:

  1. Equation of a Tangent to a Parabola (y2=4axy^2 = 4ax): For a parabola of the form y2=4axy^2 = 4ax (vertex at the origin, opening to the right), the equation of a tangent line with a slope mm is given by: y=mx+amy = mx + \frac{a}{m}

    • Why this is useful: This formula allows us to express any tangent to the parabola in terms of a single variable, its slope mm. This parametrization simplifies the process of finding a specific tangent that also satisfies other conditions.
    • Important Note: This formula is valid for m0m \ne 0. If m=0m=0, the tangent would be y=0y=0 (the x-axis), which is not a tangent to y2=4axy^2=4ax unless a=0a=0. If mm were undefined (vertical tangent, x=kx=k), the tangent to y2=4axy^2=4ax would be x=0x=0, which is the axis of the parabola and only touches it at the vertex, but cannot be expressed in y=mx+cy=mx+c form.
  2. Condition for Tangency to a Circle (x2+y2=r2x^2 + y^2 = r^2): A line Ax+By+C=0Ax + By + C = 0 is tangent to a circle x2+y2=r2x^2 + y^2 = r^2 (centered at the origin with radius rr) if and only if the perpendicular distance from the center of the circle (0,0)(0,0) to the line is equal to the radius rr. The distance formula for a point (x0,y0)(x_0, y_0) to a line Ax+By+C=0Ax+By+C=0 is Ax0+By0+CA2+B2\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}. Applying this for the center (0,0)(0,0): A(0)+B(0)+CA2+B2=r    CA2+B2=r\frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}} = r \quad \implies \quad \frac{|C|}{\sqrt{A^2 + B^2}} = r

    • Why this is useful: This condition provides a direct algebraic way to check if a given line is tangent to a circle, without needing to find the point of tangency or solve quadratic equations for intersections. It's a powerful geometric property translated into an algebraic condition.

Step-by-Step Solution

Step 1: Determine the general equation of a tangent to the parabola y2=42xy^2 = 4\sqrt{2}x.

  • Goal: Our first objective is to find the general form of any line that touches the given parabola. We will use the standard tangent formula.
  • The given parabola equation is $y^2

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