Find the points at which the graph of the equation has a vertical or horizontal tangent line. $$5 x ^ { 2 } + 4 y ^ { 2 } - 10 x + 24 y + 8 = 0$$
A) There is a vertical tangent at $y = - 3$ but no horizontal tangents.
B) There is a horizontal tangent at $x = 1$ and a vertical tangent at $y = - 3$.
C) There is a horizontal tangent at $x = 1$ but no vertical tangents.
D) There is a horizontal tangent at $x = - 2$ and a vertical tangent at $y = - 2$
E) There are no horizontal or vertical tangent lines.

Question

Quizplus · Accepted Answer

To find points with horizontal or vertical tangent lines, we differentiate the given equation implicitly with respect to $x$. The equation is $5x^2 + 4y^2 - 10x + 24y + 8 = 0$. Differentiating, we get $10x + 8yy' - 10 + 24y' = 0$. For a horizontal tangent, $y' = 0$, which simplifies to $10x - 10 = 0$, giving $x = 1$. For a vertical tangent, $y'$ is undefined, which occurs when the coefficient of $y'$ is 0, i.e., $10x - 10 = 0$, already solved for $x = 1$, but we need to solve for when the denominator of $y'$ is 0, which is a misunderstanding in the differentiation process. Correctly, for vertical tangents, we look for points where the derivative does not exist due to division by zero in the derivative formula. Solving the equation for $y$ and differentiating correctly would help identify such points, but the correct approach involves recognizing that vertical tangents occur when the derivative with respect to $y$ is zero, indicating a vertical slope. The correct differentiation and solving process reveals that there is indeed a horizontal tangent at $x = 1$ and a vertical tangent at $y = -3$, making option B correct.

Find the Points at Which the Graph of the Equation 5x2+4y2−10x+24y+8=05 x ^ { 2 } + 4 y ^ { 2 } - 10 x + 24 y + 8 = 05x2+4y2−10x+24y+8=0

Find the Points at Which the Graph of the Equation $5 x ^ { 2 } + 4 y ^ { 2 } - 10 x + 24 y + 8 = 0$