Chapter 7.CT, Problem 1CT

To determine

To match:

The equation with one of the graphs (a)-(d).

Answer to Problem 1CT

Solution:

(c).

Explanation of Solution

Given:

The equation ${4x}^2-y^2=4$.

The graphs a-d.

Concept:

Observe the given equation and check if it is of parabola, circle or ellipse.

If the given equation is quadratic for coefficient of $x$ only or coefficient of $y$ only then it is parabola.

If the given equation is quadratic for coefficients of both $x$ and $y$ plus the sign between the two is negative, then it is a hyperbola with transverse axis vertical or horizontal.

Else if the given equation is quadratic for coefficients of both $x$ and $y$ plus the sign between the two is positive, it is equation of an ellipse or circle.

For parabola, follow the following steps:-

(i) Rewrite the given equation in standard form of parabola

${\left(x-h\right)}^2=4p(y-k)$ or ${\left(y-k\right)}^2=4p(x-h)$

where the vertex is $(h,k)$.

(ii) If $p$ is positive and quadratic for coefficient of $x$ only, the parabola opens up.

If $p$ is negative and quadratic for coefficient of $x$ only, the parabola opens down.

If $p$ is positive and quadratic for coefficient of $y$ only, the parabola opens right.

If $p$ is negative and quadratic for coefficient of $y$ only, the parabola opens left.

(iii) If vertex is origin $(0, 0)$ then match the graph from options a-d having vertex as $(0, 0)$ and parabola opening up or down, right or left. Skip the further steps.

(iv) If center is not origin, then find the quadrant in which the vertex of the given equation lies.

(v) Match the quadrant found with the graph from options a-d having vertex as $(h, k)$ and parabola opening up or down, right or left.

For hyperbola, follow the following steps: -

(i) Rewrite the given equation in standard form of hyperbola with center at $(h,k)$

$\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$ or $\frac{{(y-k)}^2}{b^2}-\frac{{(x-h)}^2}{a^2}=1$.

(ii) If the given equation is of form $\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$, then it is hyperbola with transverse axis horizontal.

Else if the given equation is of form $\frac{{(y-k)}^2}{b^2}-\frac{{(x-h)}^2}{a^2}=1$, then it is hyperbola with transverse axis vertical.

(iii) Match the hyperbola with the graph with transverse axis as horizontal or vertical.

For ellipse or circle, follow the following steps: -

(i) Rewrite the given equation in standard form of ellipse with center at $(h,k)$

$\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$

and major axis length is $2a$ and minor axis length is $2b$.

(ii) If $a=b$, then it is a circle and $a=b=r$ is the radius.

If $a>b>0$, then it is an ellipse with major axis horizontal.

Else if $b>a>0$, then it is an ellipse with major axis vertical.

(iii) If center is origin $(0, 0)$ then match the graph from options a-d having center as $(0, 0)$ and major axis horizontal/vertical and length or radius as calculated. Skip the further steps.

(iv) If center is not origin, then find the quadrant in which the center point of the given equation lies.

(v) Match the quadrant found with the graph from options a-d having center as $(h, k)$ and major axis horizontal/vertical and length or radius as calculated.

Calculation:

Since given equation ${4x}^2-y^2=4$ is quadratic for coefficients of both $x$ and $y$ plus the sign between the two is negative it is a hyperbola with transverse axis vertical or horizontal.

(i) Rewrite the given equation in standard form of hyperbola with center at $(h,k)$.

$\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$ or $\frac{{(y-k)}^2}{b^2}-\frac{{(x-h)}^2}{a^2}=1$.

Hence the given equation ${4x}^2-y^2=4$ can be rewritten as

$4{(x-0)}^2-{(y-0)}^2=4$

Dividing by $4$ both sides,

$\frac{4{(x-0)}^2}{4}-\frac{{(y-0)}^2}{4}=\frac{4}{4}$

Simplifying,

$\frac{{(x-0)}^2}{1}-\frac{{(y-0)}^2}{4}=1$

$\frac{{(x-0)}^2}{1^2}-\frac{{(y-0)}^2}{2^2}=1$

The center is $(0, 0)$.

(ii) Since $\frac{{(x-0)}^2}{1^2}-\frac{{(y-0)}^2}{2^2}=1$ is of form $\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$, it is hyperbola with transverse axis horizontal.

(iii) Observing the graphs for hyperbola with center as $(0, 0)$ and transverse axis horizontal, we get the match as graph c.

Hence the graph of the equation ${4x}^2-y^2=4$ is (c).