Question

Find the equation of tangent to the curve y=5x−3−2, which is parallel to the line 4x−2y+3=0?

• A. (A) 80x−40y+103=0
• B. (B) 80x+40y+103=0
• C. (C) 80x+40y−103=0
• D. (D) 80x−40y−103=0

Answer 1

Answer: (D)

(D) 80x−40y−103=0

Answer 2

Explanation:
Given: Equation of curve is y=5x−3−2 and the tangent to the curve y=5x−3−2 is parallel to the line 4x−2y+3=0The given line 4x−2y+3=0 can be re-written as:y=2x+(32)=0Now by comparing the above equation of line with y=mx+c we get,m=2 and c=32∵ The line 4x−2y+3=0 is parallel to the tangent to the curve y=5x−3−2As we know that if two lines are parallel then their slope is same.So, the slope of the tangent to the curve y=5x−3−2 is m=2Let, the point of contact be (x1,y1)As we know that slope of the tangent at any point say (x1,y1) to a curve is given by:m=dydx⇒dydx=12⋅15x−3⋅5−0=525x−3⇒dydx=525x1−3∵ Slope of tangent to the curve y=5x−3−2 is m=2⇒2=525x1−3By squaring both the sides of the above equation we get:4=254⋅(5x1−3)⇒x1=7380∵(x1,y1) is point of conctact i.e., (x1,y1) will satisfy the equation of curve:y=5x−3−2⇒y1=5x1−3−2By substituting x1=7380 in the above equation we get:y1=−34So, the point of contact is: (7380,−34)As we know that equation of tangent at any point say (x1,y1) is given by:y−y1=dydx⋅(x−x1)⇒y+34=2⋅(x−7380)⇒80x−40y−103=0So, the equation of tangent to the given curve at the point (7380,−34) is 80x−40y−103=0Hence, the correct option is (D).