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This is one of our many quizzes on Tangent line equations. Click on this to access more such competitive quizzes on Tangents And Normals.

Tangent line equations Quiz 2

mathematics

  For a given function $f$ , we have $f\left( 3 \right) =  - 2$ and $f'\left( 3 \right) = 4$

 To solve the question, use the information given in the question to find the slope and the point of intersection. After this apply the slope-point form to find the equation of the tangent.

Given $f'\left( 3 \right) = 4$ $ \\ $ Then slope of the tangent is $4$ $ \\ $ Also, we are given that $f\left( 3 \right) =  - 2$ $ \\ $ Point of intersection is $\left( {3, - 2} \right)$ $ \\ $ Equation of the tangent line using the slope-point form $y - {y_1} = m\left( {x - {x_1}} \right)$ $ \\ y - \left( { - 2} \right) = 4\left( {x - 3} \right) \\   \Rightarrow y + 2 = 4x - 12 \\   \Rightarrow y = 4x - 14 \\  $

 If $f\left( x \right) = {x^3}$ then find the equation of the tangent line at the point where $x = 4$

 First differentiate the function to find the slope of the line. Then equate the point in the function to find the point of tangency. With the help of these find the equation of the tangent using point-slope form.

We have: $f\left( x \right) = {x^3}$ $ \\ $ On differentiating, we get: $ \\ f'\left( x \right) = 3{x^2} \\ f'\left( 4 \right) = 3{\left( 4 \right)^2} \\   \Rightarrow f'\left( 4 \right) = 48 \\  $ So, slope of the line is $48$ $ \\ $ The point of tangency: $ \\ f\left( 4 \right) = {\left( 4 \right)^3} \\   \Rightarrow f\left( 4 \right) = 64 \\  $ So, the point is $\left( {4,64} \right)$. Using the point slope form: $y - {y_1} = m\left( {x - {x_1}} \right) \\  y - 64 = 48\left( {x - 4} \right) \\   \Rightarrow y - 64 = 48x - 192 \\   \Rightarrow y = 48x - 128 \\  $

 If $y =  - {x^2} + 4x - 3$ then draw a graph of $f$ , indicating all the intercepts and turning points.

 Use the method of completing the squares to find the turning point. 

We have:  $y =  - {x^2} + 4x - 3$ $ \\ $ Complete the squares: $ \\ y =  - {x^2} + 4x - 3 \\  y =  - \left[ {{{\left( {x - 2} \right)}^2} - 4 + 3} \right] \\  y =  - {\left( {x - 2} \right)^2} + 1 \\  $ So, the point is $\left( {2,1} \right)$ $ \\ $ Now the intercepts are: $ \\ {y_i}:x = 0,y =  - 3 \\  {x_i} = y = 0 \\  $ We get, $ \\  - {x^2} + 4x - 3 = 0 \\   \Rightarrow {x^2} - 4x + 3 = 0 \\   \Rightarrow \left( {x - 3} \right)\left( {x - 1} \right) = 0 \\  $ So, $x = 1,3$ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7dffe4b0-82cc-40ba-b8e0-3db2e0bb42291504393611391185868.png"/>

 How can you find the tangent line to the parabola's equation $y = {x^2} - 2x + 7$ at $\left( {3,2} \right)$.

 First differentiate the function and then equate the point in it. After obtaining the gradient use the point-solve form to find the equation of the tangent.

We have: $y = {x^2} - 2x + 7$ $ \\ $ On differentiating both sides, we get, $ \\ \dfrac{{dy}}{{dx}} = 2x - 2 \\ $ Since, $\left( {3,2} \right)$ lies on the curve: $ \\ \dfrac{{dy}}{{dx}} = 2\left( 3 \right) - 2 \\   \Rightarrow \dfrac{{dy}}{{dx}} = 4 \\  $ Now, using $y - {y_1} = m\left( {x - {x_1}} \right)$, we get, $ \\ y - 2 = 4\left( {x - 3} \right) \\   \Rightarrow y - 2 = 4x - 12 \\   \Rightarrow y = 4x - 10 \\  $

 An ellipse is described by $16{x^2} + 9{y^2} = 85$ , find the equation of the tangent line to the ellipse at the point $x = \dfrac{1}{2}$ and $y < 0$

 First find the value of y-point of tangency by substituting the value of x in the equation. Then find the slope by differentiating the function. Use the point-slope form to find the equation.

We have: $16{x^2} + 9{y^2} = 85$ $ \\ $ Put $x = \dfrac{1}{2}$ $ \\ 16{\left( {\dfrac{1}{2}} \right)^2} + 9{y^2} = 85 \\   \Rightarrow 4 + 9{y^2} = 85 \\   \Rightarrow {y^2} = 9 \\   \Rightarrow y =  \pm 3 \\ $ We have $y < 0$, so the point is $\left( {\dfrac{1}{2}, - 3} \right)$. $ \\ $ Implicitly differentiate the above ellipse, $ \\   16\left( {2x} \right) + 9\left( {2y\dfrac{{dy}}{{dx}}} \right) = 0 \\   \Rightarrow 32x + 18y\dfrac{{dy}}{{dx}} = 0 \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 16x}}{{9y}} \\  $ So, slope of the tangent:  $ \\ f'\left( {\dfrac{1}{2}, - 3} \right) = \dfrac{{ - 16\left( {\dfrac{1}{2}} \right)}}{{9\left( { - 3} \right)}} = \dfrac{8}{{27}}$ Now, the equation of the tangent, using the point-slope form is : $y - {y_1} = m\left( {x - {x_1}} \right)$ $ \\  y - \left( { - 3} \right) = \dfrac{8}{{27}}\left( {x - \dfrac{1}{2}} \right) \\   \Rightarrow y + 3 = \dfrac{8}{{27}}\left( {x - \dfrac{1}{2}} \right) \\  $                    	   