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Question: Find the equations of tangent and normal to the ellipse \[2{x^2} + 3{y^2} = 11\] at the point whose ...

Find the equations of tangent and normal to the ellipse 2x2+3y2=112{x^2} + 3{y^2} = 11 at the point whose ordinate is 1.

Explanation

Solution

We use the given value of ordinate and substitute in the equation of the ellipse to calculate the abscissa at that coordinate. Differentiate the equation of ellipse with respect to x and find the value of tangent and normal using the formula. Substitute the values of x and y in the equation of tangent and normal obtained in the starting.

  • y=f(x)y = f\left( x \right) is a function with a point A(x1,y1){\text{A}}\left( {{x_1},{y_1}} \right), then, the slope of the tangent is (dydx)x=x1;y=y1{\left( {\dfrac{{dy}}{{dx}}} \right)_{x = {x_1};y = {y_1}}} and the slope of the normal is −1(dydx)x=x1;y=y1\dfrac{{ - 1}}{{{{\left( {\dfrac{{dy}}{{dx}}} \right)}_{x = {x_1};y = {y_1}}}}}, where dydx\dfrac{{dy}}{{dx}} is known as the implicit differentiation w.r.t to xx.

Complete step-by-step solution:
We are given the equation as 2x2+3y2=112{x^2} + 3{y^2} = 11
For calculating points of intersection at y=1y = 1, we will substitute this value in the equation 2x2+3y2=112{x^2} + 3{y^2} = 11
⇒2x2+3(1)2=11\Rightarrow 2{x^2} + 3{\left( 1 \right)^2} = 11
By simplifying the above expression, we get:
⇒2x2+3=11\Rightarrow 2{x^2} + 3 = 11
Bringing 33 into the RHS side of the above equation and subtracting it from 1111, we get:
⇒2x2=8\Rightarrow 2{x^2} = 8
Bringing 22 into the RHS side of the above equation and dividing it from 88, we get:
⇒x2=4\Rightarrow {x^2} = 4
Taking root on both the sides of the above equation we get the two values xx as given below:
⇒x=±2\Rightarrow x = \pm 2
So, at y=1y = 1, the points are (2,1)\left( {2,1} \right) and (−2,1)\left( { - 2,1} \right).
Now we will differentiate the equation given in the question with respect to x.
By doing the implicit differentiation w.r.txx in the equation 2x2+3y2=112{x^2} + 3{y^2} = 11, we get:
⇒4x+6y(dydx)=0\Rightarrow 4x + 6y\left( {\dfrac{{dy}}{{dx}}} \right) = 0
Bringing 4x4x into the RHS side of the above equation, we get:
⇒6y(dydx)=−4x\Rightarrow 6y\left( {\dfrac{{dy}}{{dx}}} \right) = - 4x
Bringing 6y6y into the RHS side of the above equation and dividing it from −4x - 4x, we get:
⇒(dydx)=−4x6y\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{ - 4x}}{{6y}}
Calculate the slope of the tangent at the point (2,1)\left( {2,1} \right)using the formula of slope of tangent at a point i.e. (dydx)x=x1;y=y1{\left( {\dfrac{{dy}}{{dx}}} \right)_{x = {x_1};y = {y_1}}}
⇒−4(2)6(1)=−43\Rightarrow \dfrac{{ - 4\left( 2 \right)}}{{6\left( 1 \right)}} = \dfrac{{ - 4}}{3}
Calculate the slope of the normal at the point (2,1)\left( {2,1} \right)using the formula of slope of normal at point i.e. −1(dydx)x=x1;y=y1\dfrac{{ - 1}}{{{{\left( {\dfrac{{dy}}{{dx}}} \right)}_{x = {x_1};y = {y_1}}}}}
⇒6(1)4(2)=34\Rightarrow \dfrac{{6\left( 1 \right)}}{{4\left( 2 \right)}} = \dfrac{3}{4}
Now calculate the equation of the tangent at the point (2,1)\left( {2,1} \right) using the formula m=y−y1x−x1m = \dfrac{{y - {y_1}}}{{x - {x_1}}}
⇒y−1=m(x−2)\Rightarrow y - 1 = m\left( {x - 2} \right)
By substituting the value of slope m=−43m = \dfrac{{ - 4}}{3} into the above equation we get:
⇒y−1=−43(x−2)\Rightarrow y - 1 = \dfrac{{ - 4}}{3}\left( {x - 2} \right)
Bringing 33 into the LHS side of the above equation and multiplying it with y−1y - 1, we get:
⇒3y−3=−4(x−2)\Rightarrow 3y - 3 = - 4\left( {x - 2} \right)
By doing the multiplication in the RHS side of the above equation we get:
⇒3y−3=−4x+8\Rightarrow 3y - 3 = - 4x + 8
Bringing xx and yy terms into the LHS side of the equation we get:
⇒4x+3y=11\Rightarrow 4x + 3y = 11
Now calculate the equation of the normal at the point (2,1)\left( {2,1} \right) using the formula m=y−y1x−x1m = \dfrac{{y - {y_1}}}{{x - {x_1}}}
⇒y−1=m(x−2)\Rightarrow y - 1 = m\left( {x - 2} \right)
By substituting the value of slope m=34m = \dfrac{3}{4} into the above equation we get:
⇒y−1=34(x−2)\Rightarrow y - 1 = \dfrac{3}{4}\left( {x - 2} \right)
Bringing 44 into the LHS side of the above equation and multiplying it with y−1y - 1, we get:
⇒4y−4=3(x−2)\Rightarrow 4y - 4 = 3\left( {x - 2} \right)
By doing the multiplication in the RHS side of the above equation we get:
⇒4y−4=3x−6\Rightarrow 4y - 4 = 3x - 6
Bringing xx and yy terms into the LHS side of the equation we get:
⇒4y−3x=−2\Rightarrow 4y - 3x = - 2
Now we calculate the slope of the tangent at the point (−2,1)\left( { - 2,1} \right) using the formula of slope of tangent at a point i.e. (dydx)x=x1;y=y1{\left( {\dfrac{{dy}}{{dx}}} \right)_{x = {x_1};y = {y_1}}}
⇒−4(−2)6(1)=43\Rightarrow \dfrac{{ - 4\left( { - 2} \right)}}{{6\left( 1 \right)}} = \dfrac{4}{3}
Calculate the slope of the normal at the point (−2,1)\left( { - 2,1} \right) using the formula of slope of normal at a point i.e. −1(dydx)x=x1;y=y1\dfrac{{ - 1}}{{{{\left( {\dfrac{{dy}}{{dx}}} \right)}_{x = {x_1};y = {y_1}}}}}
⇒6(1)4(−2)=−34\Rightarrow \dfrac{{6\left( 1 \right)}}{{4\left( { - 2} \right)}} = - \dfrac{3}{4}
Now calculate the equation of the tangent at the point (−2,1)\left( { - 2,1} \right) using the formula m=y−y1x−x1m = \dfrac{{y - {y_1}}}{{x - {x_1}}}
⇒y−1=m(x+2)\Rightarrow y - 1 = m\left( {x + 2} \right)
By substituting the value of slope m=43m = \dfrac{4}{3} into the above equation we get:
⇒y−1=43(x+2)\Rightarrow y - 1 = \dfrac{4}{3}\left( {x + 2} \right)
Bringing 33 into the LHS side of the above equation and multiplying it with y−1y - 1, we get:
⇒3y−3=4(x+2)\Rightarrow 3y - 3 = 4\left( {x + 2} \right)
By doing the multiplication in the RHS side of the above equation we get:
⇒3y−3=4x+8\Rightarrow 3y - 3 = 4x + 8
Bringing xx and yy terms into the LHS side of the equation we get:
⇒3y−4x=11\Rightarrow 3y - 4x = 11
Now calculate the equation of the normal at the point (−2,1)\left( { - 2,1} \right) using the formula m=y−y1x−x1m = \dfrac{{y - {y_1}}}{{x - {x_1}}}
⇒y−1=m(x+2)\Rightarrow y - 1 = m\left( {x + 2} \right)
By substituting the value of slope m=−34m = \dfrac{{ - 3}}{4} into the above equation we get:
⇒y−1=−34(x+2)\Rightarrow y - 1 = \dfrac{{ - 3}}{4}\left( {x + 2} \right)
Bringing 44 into the LHS side of the above equation and multiplying it with y−1y - 1, we get:
⇒4y−4=−3(x+2)\Rightarrow 4y - 4 = - 3\left( {x + 2} \right)
By doing the multiplication in the RHS side of the above equation we get:
⇒4y−4=−3x−6\Rightarrow 4y - 4 = - 3x - 6
Bringing xx and yy terms into the LHS side of the equation we get:
⇒3x+4y=−2\Rightarrow 3x + 4y = - 2
∴\therefore The equations of tangents are 4x+3y=114x + 3y = 11 and 3y−4x=113y - 4x = 11. And the equation of slope is 3x+4y=−23x + 4y = - 2 and 4y−3x=−24y - 3x = - 2

Note: Many students make the mistake of only taking the value of abscissa as 2 and ignore the case of -2. Keep in mind whenever we remove under root of a perfect square we get both negative and positive values of the number. Also, when solving for the equation of tangent and normal, many students tend to leave the equation in ‘x’ and ‘y’ terms i.e. they don’t substitute the obtained values of abscissa and ordinate. Keep in mind we have to find value at that particular point where ordinate is 1 so we have to put in value of the point.