Solve for the equation in standard form of the following conic sections and graph the curve on a Cartesian plane indicating important points. 1. An ellipse passing through (4, 3) and (6, 2) 2. A parabola with axis parallel to the x-axis and passing through (5, 4), (11, -2) and (21, -4) 3. The hyperbola given by 5x2−4y2=20x+24y+36.

Question

Solve for the equation in standard form of the following conic sections and graph the curve on a Cartesian plane indicating important points.  1. An ellipse passing through (4, 3) and (6, 2)  2. A parabola with axis parallel to the x-axis and passing through (5, 4), (11, -2) and (21, -4)  3. The hyperbola given by 5x2−4y2=20x+24y+36.

Ezra Herbert · Accepted Answer

Step 1  An equation of the ellipse is of the form,  x2a2+y2b2=1  Equation of the ellipse through the point (4, 3).  So, x=4,y=3 satisfies the equation of an ellipse.  Implies that,  42a2+32b2=1  ⇒16a2+9b2=1  ⇒16a2=1−9b2  ⇒1a2=116(1−9b2)  Step 2  Equation of the ellipse through the point (6, 2).  So, x=6,y=2 satisfies the equation of an ellipse.  ⇒62a2+22b2=1  ⇒36a2+4b2=1  ⇒36a2=1−4b2  ⇒1a2=136(1−4b2)  Step 3  Compare equations (1) and (2), we get  136(1−4b2)=116(1−9b2)  Multiply both sides by 4,  19(1−4b2)=14(1−9b2)  Distribute both sides,  19−49b2=14−94b2  Combine like terms  19−14=49b2−94b2  Make the same denominator,  436−936=1636b2−8136b2  ⇒−536=−6536b2  ⇒−536=−6536b2  Multiply both sides by 36,  ⇒−51=−65b2  Divide both sides by -65,  ⇒565=1b2  By simplifying it,  ⇒113=1b2  Taking reciprocal from both sides,  ⇒b2=13  Step 4  Substitute b2=13   in the equation 1a2=116(1−9b2),

Solve for the equation in standard form of the following conic sections and graph the curve on a Cartesian plane indicating important points. 1. An el

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