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Number of Values of c for Which y = cx + c Touches an Ellipse | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question

For how many values of

c

does the line

y = cx + c

c \in \mathbb{R}

, touch the curve

\dfrac{x^{2}}{4} + \dfrac{y^{2}}{1} = 1

0

correct

1

2

Dmore than

2

Solution

Step 1: In

y = cx + c

the slope and the intercept are the same number

c

, so this is a one-parameter family of lines, not the usual two-parameter family. Step 2: The line

y = mx + k

touches

\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1

exactly when

k^{2} = a^{2}m^{2} + b^{2}

Here

a^{2}=4

b^{2}=1

m = c

and

k = c

. Step 3:

c^{2} = 4c^{2} + 1 \ \Longrightarrow\ 3c^{2} + 1 = 0 \ \Longrightarrow\ c^{2} = -\frac13

No real

c

satisfies this, so there is no such tangent line at all. Step 4: Every line of the family passes through

(-1,0)

, and

\dfrac{1}{4} + 0 = \dfrac14 < 1

, so

(-1,0)

lies inside the ellipse. A line through an interior point can never be a tangent, which confirms the count of zero. Answer: (1)

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