Respuesta :
The slope of the tangent at point [tex]P = \left( {1, - 2} \right)[/tex] is [tex]\boxed{\frac{2}{3}}.[/tex]
Further explanation:
Given:
The equation of the ellipse is [tex]2{x^2} + xy + {y^2} = 4.[/tex]
The point [tex]P = \left( {1, - 2} \right)[/tex] is on the ellipse and the tangent.
Explanation:
The given equation of the ellipse is [tex]2{x^2} + xy + y = 4.[/tex]
Now differentiate the equation of ellipse with respect to [tex]x[/tex].
[tex]\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}[/tex]
Further solve the above equation.
[tex]\begin{aligned}\left( {x + 2y}\right)\frac{{dy}}{{dx}}&=- 4x - y\\\frac{{dy}}{{dx}}&=-\frac{{4x + y}}{{\left( {x + 2y} \right)}}\\\end{aligned}[/tex]
Substitute [tex]1[/tex] for [tex]x[/tex] and [tex]-2[/tex] for y in equation [tex]\dfrac{{dy}}{{dx}}= - \dfrac{{4x + y}}{{x + 2y}}.[/tex]
[tex]\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}[/tex]
The slope of the tangent at point [tex]P = \left( {1, - 2} \right)[/tex] is [tex]\boxed{\frac{2}{3}}.[/tex]
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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.
The slope of the tangent line through the point P(1, -2) is;
Slope = 2/3
We are given the equation of an ellipse as;
2x² + xy + y² = 4
Now, in mathematics we know that the slope of a tangent is also the same as the first derivative. Thus, let us differentiate the equation of the ellipse with respect to x. Thus;
4x(dx) + x(dy) + y(dx) + 2y(dy) = 0
Collecting like terms;
(4x + y)dx + (x + 2y)dy = 0
⇒ (4x + y)dx = - (x + 2y)dy
divide both sides by dx to get;
(4x + y) = - (x + 2y)(dy/dx)
dy/dx = (4x + y)/-(x + 2y)
At point P(1, -2), the slope is;
dy/dx = (4(1) + -2)/-(1 + 2(-2))
dy/dx = 2/3
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