# How to find equations of both lines that are tangent to the curve y = 1 + x^2 and are parallel to the line 12x - y=1?

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Parallel lines have an equal slope. As the equation of the required tangents to the curve are parallel to the line 12x - y = 1, the slope of the tangents is equal to the slope of 12x - y = 1.

12x - y = 1

=> y = 12x - 1

This is the equation of the line in the slope-intercept form with the slope given by 12.

The value of the first derivative at any point on the curve is equal to the slope of the tangent at that point.

For the curve y = 1 + x^2, y' = 2x

2x = 12

=> x = 6

For x = 6, y = 1 + 6^2 = 37

The required tangent passes through (6, 37) and has a slope 12.

(y - 37)/(x - 6) = 12

= y - 37 = 12x - 72

=> 12x - y - 35 = 0

There is only one tangent to the curve that is parallel to the line 12x - y = 1.

**The equation of the tangent is : 12x - y - 35 = 0**