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Visual SolutionStandard

Question

If the line

y = mx + c

is a tangent to the parabola

y^2 = 4x

, then the relation between

m

and

c

is:

(A)

c = \frac{1}{m}

(B)

c = m

(C)

c = \frac{m}{2}

(D)

c = \frac{2}{m}

Solution Path

Substitute line into parabola, discriminant = 0 gives

c = 1/m

01Question Setup

1/4

Find the condition for

y = mx + c

to be tangent to

y^2 = 4x

Tangent condition

= ;?

02Plot Parabola

2/4

Draw

y^2 = 4x

with focus at

(1, 0)

and directrix

x = -1

y^2 = 4x

, focus

(1, 0)

03Tangent ConditionKEY INSIGHT

3/4

Substitute

y = mx + c

into

y^2 = 4x

. Set discriminant

= 0

(mc)^2 - 4(mc - 1) \cdot m = 0

, giving

c = 1/m

c = \frac{1}{m}

04Final Answer

4/4

The tangent to

y^2 = 4ax

in slope form is

y = mx + a/m

. Here

a = 1

, so

c = 1/m

\boxed{c = \frac{1}{m}}

Concepts from this question2 concepts unlocked

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Tangent Condition for Conics

STANDARD

For the line y = mx + c to be tangent to the ellipse x²/a² + y²/b² = 1, the condition is c² = a²m² + b². Similar conditions exist for parabola (c = a/m) and hyperbola (c² = a²m² - b²).

c^2 = a^2 m^2 + b^2 \quad \text{(ellipse)}

One of the most tested ideas in JEE. Lets you find tangent equations, common tangents, and locus of intersection points without calculus.

Common tangentsTangent from external pointLocus of tangent intersection

Practice (14 Qs) →

Eccentricity and Conic Classification

EASY

The eccentricity e = c/a determines the type of conic: e < 1 gives an ellipse, e = 1 gives a parabola, and e > 1 gives a hyperbola. For a circle, e = 0.

e = \frac{c}{a}

First step in most conic problems. JEE frequently asks you to identify or compare conics based on eccentricity values.

Conic identificationEccentricity comparisonLocus problems

Practice (12 Qs) →

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