# There are two distinct straight lines that pass throgh the point (1,-3) and are tangent to the curve y=x^2. Find their equations.

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Expert Answers

justaguide | Certified Educator

The lines passing through the point (1, -3) are tangent to the curve y = x^2. We have to find their equations.

The slope of the line tangential to the curve y = x^2 is given by 2x

The tangent touches the curve at (x, x^2) and (1, -3)

This gives (x^2 + 3)/(x - 1) = 2x

=> x^2 + 3 = 2x^2 - 2x

=> x^2 - 2x - 3 = 0

=> x^2 - 3x + x - 3 = 0

=> x(x - 3) + 1(x - 3) = 0

=> (x + 1)(x - 3) = 0

=> x = -1 and x = 3

y = 1 and 9

The equation of the tangents are (y - 9)/(x - 3) = 6

=> y - 9 = 6x - 18

=> 6x - y - 9 = 0

and

(y - 1)/(x + 1) = -2

=> y - 1 = -2x - 2

=> 2x + y + 1 = 0

**The required equations of the tangents is 6x - y - 9 = 0 and 2x + y + 1 = 0**