The slope of the tangent at point \(P = \left( {1, - 2} \right)\) is \(\boxed{\frac{2}{3}}.\)
Further explanation:
Given:
The equation of the ellipse is \(2{x^2} + xy + {y^2} = 4.\)
The point \(P = \left( {1, - 2} \right)\) is on the ellipse and the tangent.
Explanation:
The given equation of the ellipse is \(2{x^2} + xy + y = 4.\)
Now differentiate the equation of ellipse with respect to \(x\).
\(\begin{aligned}\frac{d}{{dx}}\left( {2{x^2} + xy + {y^2}} \right) &= \frac{d}{{dx}}\left( 4 \right)\\\frac{d}{{dx}}\left( {2{x^2}} \right) + \frac{d}{{dx}}\left( {xy} \right) + \frac{d}{{dx}}\left( {{y^2}} \right)&= \frac{d}{{dx}}\left( 4 \right)\\4x + x \times \frac{{dy}}{{dx}} + y\frac{d}{{dx}}\left( x \right) + 2y\frac{{dy}}{{dx}}&= 0\\4x + y + \left( {x + 2y} \right)\frac{{dy}}{{dx}}&= 0\\4x + y + \left( {x + 2y} \right)\frac{{dy}}{{dx}} &= 0 \\\end{aligned}\)
Further solve the above equation.
\(\begin{aligned}\left( {x + 2y}\right)\frac{{dy}}{{dx}}&=- 4x - y\\\frac{{dy}}{{dx}}&=-\frac{{4x + y}}{{\left( {x + 2y} \right)}}\\\end{aligned}\)
Substitute \(1\) for \(x\) and \(-2\) for y in equation \(\dfrac{{dy}}{{dx}}= - \dfrac{{4x + y}}{{x + 2y}}.\)
\(\begin{aligned}\frac{{dy}}{{dx}} &= - \frac{{4\left( 1 \right) + \left( { - 2} \right)}}{{1 + 2\left( { - 2} \right)}} \\&= - \frac{{4 - 2}}{{1 - 4}}\\&= - \frac{2}{{ - 3}}\\&= \frac{2}{3} \\\end{aligned}\)
The slope of the tangent at point \(P = \left( {1, - 2} \right)\) is \(\boxed{\frac{2}{3}}.\)
Learn more:
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2. Learn more about equation of circle brainly.com/question/1506955.
3. Learn more about range and domain of the function https://brainly.com/question/3412497
Answer details:
Grade: High School
Subject: Mathematics
Chapter: Application of derivatives
Keywords: Derivative, graph of the equation, 2x2+xy+y2=4, ellipse, tangent, equation yield, pictured ellipse, line, intersect, slope, normal, slope of tangent line.
